3. Partial Fractions

Exercises

a. General Decompositions

Write out the general partial fraction expansion for each fraction. Do not find the numerical values of the coefficients. You will find the coefficients in the exercises for the next section, below.

\(\dfrac{12}{x^2+x-2}\)

\(x^2+x-2=(x+2)(x-1)\)

\(\dfrac{12}{x^2+x-2}=\dfrac{A}{x+2}+\dfrac{B}{x-1}\)

We factor the denominator: \(x^2+x-2=(x+2)(x-1)\). Since each factor is linear, we put constants \(A\) and \(B\) over each factors: \[ \dfrac{12}{x^2+x-2}=\dfrac{A}{x+2}+\dfrac{B}{x-1} \]

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\(\dfrac{4x^2-6}{x^3-x^2-6x}\)

First factor an \(x\) out of the denominator.

\(\dfrac{4x^2-6}{x^3-x^2-6x}=\dfrac{A}{x}+\dfrac{B}{x+2}+\dfrac{C}{x-3}\)

Factor out a common \(x\). Now factor \(x^2-x-6\): \((x+2)(x-3)\). Now put letters over the factors. Don't forget \(x\): \(\dfrac{A}{x}+\dfrac{B}{x+2}+\dfrac{C}{x-3}\)

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\(\dfrac{1}{x^4-1}\)

Difference of squares. Twice!

\(\dfrac{1}{x^4-1}=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x+1}+ \dfrac{D}{x-1}\)

\(\dfrac{1}{x^4-1}\) is a difference of two squares: \((x^2+1)(x^2-1)\).

\(x^2-1\) can still be factored: \((x+1)(x-1)\). Now put letters on top. Remember that since

\(x^2+1\) has a 2nd degree and cannot be factored, it will have an \(Ax+B\) on top:

\(\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x+1}+ \dfrac{D}{x-1}\)

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\(\dfrac{9}{x^3+9x}\)

Factor an \(x\) out of the denominator.

\(\dfrac{9}{x^3+9x}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+9}\)

Factor out the \(x\) first. \(x^2+9\) cannot be factored further so it is left alone. Remember that \(x^2+9\) has a 2nd degree: \(\dfrac{A}{x}+\dfrac{Bx+C}{x^2+9}\)

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\(\dfrac{2x^2+1}{x^2+2x}\)

Factor an \(x\) out of the denominator.

\(\dfrac{2x^2+1}{x^2+2x}=2+\dfrac{A}{x}+\dfrac{B}{x+2}\)

Since the top and bottom have the same degree, we must divide. After dividing, we get \(2\) with a remainder of \(-4x+1\). However, this is still over \(x^2+2x\). We then factor to get \((x)(x+2)\). Putting all this together: \(2+\dfrac{A}{x}+\dfrac{B}{x+2}\)

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\(\dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)}\)

\(\dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)} =\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}+\dfrac{Cx+D}{x^2+1}\)

It is easy enough to see the factors: \((x+2)^2\) and \(x^2+1\). However, \((x+2)^2\) must be written as \((x+2)\) and \((x+2)^2\) because ......? Also, since \(x^2+1\) cannot be factored, there will be a linear numerator: \(\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}+\dfrac{Cx+D}{x^2+1}\)

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\(\dfrac{2x^3-x^2}{(x^2+1)^2}\)

\(\dfrac{2x^3-x^2}{(x^2+1)^2} =\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{(x^2+1)^2}\)

Similar to the last one, this one must be written as \(x^2+1\) and \((x^2+1)^2\). However, both of them cannot be factored so they are both written as linear terms: \(\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{(x^2+1)^2}\)

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b. Finding Coefficients

Find the coefficients in the partial fraction expansion for each fraction. The general expansion was found in the previous section.

\(\dfrac{12}{x^2+x-2}\)
See a previous problem.

\(\dfrac{12}{x^2+x-2}=\dfrac{-4}{x+2}+\dfrac{4}{x-1}\)

We clear the denominator: \[ 12=(x-1)A+(x+2)B \] Plugging in \(1\): \[ \quad 12=3B\] \[ \quad 4=B \] Plugging in \(-2\): \[ \quad 12=-3A \] \[ \quad -4=A \] Putting these two together we get: \[ \dfrac{-4}{x+2}+\dfrac{4}{x-1} \]

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\(\dfrac{4x^2-6}{x^3-x^2-6x}\)
See a previous problem.

\(\dfrac{4x^2-6}{x^3-x^2-6x}=\dfrac{1}{x}+\dfrac{1}{x+2}+\dfrac{2}{x-3}\)

We first expand: \[ \dfrac{4x^2-6}{x^3-x^2-6x} =\dfrac{A}{x}+\dfrac{B}{x+2}+\dfrac{C}{x-3} \] We clear the denominator: \[4x^2-6=(x+2)(x-3)A+(x)(x-3)B+(x)(x+2)C \] Plugging in \(0\): \[-6=-6A\] \[1=A\] Plugging in \(-2\): \[10=10B\] \[1=B\] Plugging in \(3\): \[30=15C\] \[2=C\] Finally, we end up with: \[ \dfrac{4x^2-6}{x^3-x^2-6x} =\dfrac{1}{x}+\dfrac{1}{x+2}+\dfrac{2}{x-3} \]

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\(\dfrac{1}{x^4-1}\)
See a previous problem.

\(\dfrac{1}{x^4-1} =\dfrac{1}{4}\left[-\dfrac{2}{x^2+1}-\dfrac{1}{x+1}+\dfrac{1}{x-1}\right]\)

We clear the denominator: \[ 1=(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1) \] Plugging in \(1\): \[ \quad 1=D(2)(2)\] \[ \quad 1=4D\] \[ \quad \dfrac{1}{4}=D \] Plugging in \(-1\): \[\quad 1=C(2)(-2) \] \[ \quad 1=-4C \] \[ \quad \dfrac{-1}{4}=C \] Plugging in \(0\): \[\quad 1=-B-C+D \] Substituting \(D\) and \(C\): \[\quad 1=-B+\dfrac{1}{4}+\dfrac{1}{4} \] \[\quad \dfrac{1}{2}=-B \] \[\quad \dfrac{-1}{2}=B \] Plugging in \(2\): \[ \quad 1=(2A+B)(3)(1)+C(5)(1)+D(5)(3) \] \[ \quad 1=6A+3B+5C+15D \] Substituting \(B\), \(C\), and \(D\): \[ \quad 1=6A-\dfrac{3}{2}-\dfrac{5}{4}+\dfrac{15}{4} \] \[ \quad 1=6A+1\] \[ \quad 0=6A \] \[ \quad 0=A \] Putting these all together: \[ \dfrac{-1/2}{x^2+1}-\dfrac{1/4}{x^2+1}+\dfrac{1/4}{x^2+1} \] We factor out a \(\dfrac{1}{4}\) to get: \[ \dfrac{1}{4}\left[-\dfrac{2}{x^2+1}-\dfrac{1}{x+1}+ \dfrac{1}{x-1}\right] \]

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\(\dfrac{9}{x^3+9x}\)
See a previous problem.

\(\dfrac{9}{x^3+9x}=\dfrac{1}{x}-\dfrac{x}{x^2+9}\)

We clear the denominator: \[ 9=(x^2+9)A+(x)(Bx+C) \] Plugging in \(0\): \[ \quad 9=9A \] \[ \quad 1=A \] Plugging in \(1\): \[ \quad 9=10A+B+C \] Substituting \(A\): \[ 9=10+B+C \] \[-1=B+C\] Plugging in \(2\): \[ \quad 9=13A+4B+2C \] Substituting \(A\): \[ \quad 9=13+4B+2C \] \[\quad -4=4B+2C \] \[\quad -4=2B+2C+2B\] Substituting the 2nd equations: \[ \quad -4=-2+2B\] \[\quad -2=2B\] \[\quad -1=B\] Substituting \(B\) into the 2nd equation: \[ \quad -1=-1+C\] \[ \qquad 0=C\] Finally, we get: \[ \dfrac{9}{x^3+9x}=\dfrac{1}{x}-\dfrac{x}{x^2+9} \]

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\(\dfrac{2x^2+1}{x^2+2x}\)
See a previous problem.

\(\dfrac{2x^2+1}{x^2+2x}=2+\dfrac{1}{2x}-\dfrac{9}{2(x+2) }\)

We clear the denominator: \[ 4x-1=A(x+2)+Bx=(A+B)x+2A \] Equating coefficients yields \[ A+B=4\qquad and\qquad 2A=-1 \] So \[ A=-1/2\qquad and\qquad B=9/2 \] Putting all this together gives \[ \dfrac{2x^2+1}{x^2+2x}=2-\left(\dfrac{-1/2}{x}+\dfrac{9/2}{x+2}\right) =2+\dfrac{1}{2x}-\dfrac{9}{2(x+2) } \]

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\(\dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)}\)
See a previous problem.

\(\dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)}=\dfrac{-1}{x+2}+\dfrac{2}{(x+2)^2}+ \dfrac{x+1}{x^2+1}\)

We clear the denominator \[ 5x^2+7x+4=(x+2)(x^2+1)A+(x^2+1)B+(x+2)^2(Cx+D) \] Substituting \(-2\): \[20-14+4=5B \] \[ 10=5B \] \[ 2=B \] Substituting \(0\): \[ 4=2A+B+4D \] Substituting \(B=2\): \[ 4=2A+2+4D\] \[ 2=2A+4D\] \[ 1=A+2D \qquad (Eq.1)\] Substituting \(1\): \[5+7+4=6A+2B+9C+9D \] Substituting \(B=2\): \[ 16=6A+4+9C+9D \] \[ 12=6A+9C+D \qquad (Eq.2)\] Substituting \(-1\): \[5-7+4=2A+2B-C+D \] Substituting \(B=2\): \[ 2=2A+4-C+D \] \[-2=2A-C+D \qquad (Eq.3)\] Multiplying \(Eq.3\) by \(9\): \[-18=18A-9C+9D \qquad (Eq.4)\] Adding \(Eq.2\) and \(Eq.4\): \[-6=24A+18D \] \[-6=(9A+18D)+15A \] Substituting \(Eq. 1\): \[-6=9+15A \] \[-15=15A \] \[-1=A \] Substituting the \(A\) value into \(Eq.1\): \[ 1=-1+2D \] \[ 2=2D \] \[ 1=D \] Substituting \(A\) and \(D\) into \(Eq. 3\): \[-2=-2-C+1 \] \[-1=-C \] \[ 1=C \] Putting this all together gives us: \[ \dfrac{-1}{x+2}+\dfrac{2}{(x+2)^2}+ \dfrac{x+1}{x^2+1} \]

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\(\dfrac{2x^3-x^2}{(x^2+1)^2}\)
See a previous problem.

\(\dfrac{2x^3-x^2}{(x^2+1)^2}=\dfrac{2x-1}{x^2+1}+\dfrac{-2x+1}{(x^2+1)^2}\)

We clear the denominator: \[2x^3-x^2=(Ax+B)(x^2+1)+Cx+D \] Plugging in \(0\): \[ 0=B+D \qquad (Eq.1)\] Plugging in \(1\) and \(-1\): \[1: 1=2A+2B+C+D \qquad (Eq.2)\] \[-1:-3=-2A+2B-C+D \] Adding these two equations: \[-2=4B+2D \qquad (Eq.3) \] \[-2=(2B+2D)+2B \] Substituting \(Eq.1\): \[-2=2B \] \[-1=B \] Substituting \(B\) into \(Eq.3\) we see that \[ D=1 \] We now plug in \(2\) to the original equation: \[ 12=10A+5B+2C+D \] Substituting the B and D values: \[ 12=10A-5+2C+1 \] \[ 16=10A+2C \qquad (Eq.4) \] We now look back to Eq.2 and substitute B and D: \[ 1=2A-2+C+1 \] \[ 2=2A+C \qquad (Eq.5)\] We multiply \(Eq.5\) by 2: \[ 4=4A+2C \qquad (Eq.6) \] We then subtract \(Eq.6\) from \(Eq.4\): \[ 12=6A \] \[ 2=A \] Substituting \(A\) into \(Eq.5\): \[ 2=4+C \] \[-2=C \] Putting this all together and we get: \[ \dfrac{2x-1}{x^2+1}+\dfrac{-2x+1}{(x^2+1)^2} \]

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c. Computing Integrals

Compute each integral.

\(\displaystyle \int \dfrac{12}{x^2+x-2}\,dx\)
See a previous problem.

From a previous problem, \[ \dfrac{12}{x^2+x-2}=\dfrac{-4}{x+2}+\dfrac{4}{x-1} \]

\(\displaystyle \int \dfrac{12}{x^2+x-2}\,dx =-4\ln |x+2|+4\ln |x-1|+C\)

From a previous problem, \[ \dfrac{12}{x^2+x-2}=\dfrac{-4}{x+2}+\dfrac{4}{x-1} \] So: \[ \begin{aligned} \int &\dfrac{12}{x^2+x-2}\,dx = \int \left( \dfrac{-4}{x+2}+\dfrac{4}{x-1} \right)\,dx \\ &= -4\ln |x+2|+4\ln |x-1|+C \end{aligned} \]

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\(\displaystyle \int \dfrac{4x^2-6}{x^3-x^2-6x}\,dx\)
See a previous problem.

From a previous problem, \[ \dfrac{4x^2-6}{x^3-x^2-6x}=\dfrac{1}{x}+\dfrac{1}{x+2}+\dfrac{2}{x-3} \]

\(\displaystyle\ln |x|+\ln |x+2|+2\ln |x-3|+C\)

From a previous problem, \[ \dfrac{4x^2-6}{x^3-x^2-6x}=\dfrac{1}{x}+\dfrac{1}{x+2}+\dfrac{2}{x-3} \] So: \[ \begin{aligned} \int &\dfrac{4x^2-6}{x^3-x^2-6x} =\int\left(\dfrac{1}{x}+\dfrac{1}{x+2}+\dfrac{2}{x-3}\right)\,dx\\ &=\ln |x|+\ln |x+2|+2\ln |x-3|+C \end{aligned} \]

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\(\displaystyle \int \dfrac{1}{x^4-1}\,dx\)
See a previous problem.

From a previous problem, \[ \dfrac{1}{x^4-1}=\dfrac{1}{4}\left[-\dfrac{2}{x^2+1}-\dfrac{1}{x+1}+\dfrac{1}{x-1}\right] \]

\(\displaystyle \int\dfrac{1}{x^4-1} =\dfrac{1}{4}\left[\rule{0pt}{10pt}-2\arctan(x)+\ln\left|\dfrac{x-1}{x+1}\right|\right]+C\)

From a previous problem, \[ \dfrac{1}{x^4-1}=\dfrac{1}{4}\left[-\dfrac{2}{x^2+1}-\dfrac{1}{x+1}+\dfrac{1}{x-1}\right] \] So: \[ \begin{aligned} \int &\dfrac{1}{x^4-1} =\int\dfrac{1}{4}\left[-\dfrac{2}{x^2+1}-\dfrac{1}{x+1}+\dfrac{1}{x-1}\right]\,dx\\ &=\dfrac{1}{4}\left[\rule{0pt}{10pt}-2\arctan(x)-\ln|x+1|+\ln|x-1|\right]+C \\ &=\dfrac{1}{4}\left[\rule{0pt}{10pt}-2\arctan(x)+\ln\left|\dfrac{x-1}{x+1}\right|\right]+C \end{aligned} \]

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\(\displaystyle \int \dfrac{9}{x^3+9x}\,dx\)
See a previous problem.

From a previous problem, \[ \dfrac{9}{x^3+9x}=\dfrac{1}{x}-\dfrac{x}{x^2+9} \]

\(\displaystyle \int \dfrac{9}{x^3+9x}\,dx =\ln |x|-\dfrac{1}{2}\ln(x^2+9)+C\)

From a previous problem, \[ \dfrac{9}{x^3+9x}=\dfrac{1}{x}-\dfrac{x}{x^2+9} \] Then: \[ \begin{aligned} \int \dfrac{9}{x^3+9x}\,dx &= \int \left(\dfrac{1}{x}-\dfrac{x}{x^2+9}\right) \,dx \\ &= \ln |x|-\dfrac{1}{2}\ln |x^2+9|+C \end{aligned} \]

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\(\displaystyle \int \dfrac{2x^2+1}{x^2+2x}\,dx\)
See a previous problem.

From a previous problem, \[ \dfrac{2x^2+1}{x^2+2x} =2+\dfrac{1}{2x}-\dfrac{9}{2(x+2)} \]

\(\displaystyle \int \dfrac{2x^2+1}{x^2+2x}\,dx =2x+\dfrac{1}{2}\ln |x|-\dfrac{9}{2}\ln |x+2|+C\)

From a previous problem, \[ \dfrac{2x^2+1}{x^2+2x} =2+\dfrac{1}{2x}-\dfrac{9}{2(x+2)} \] Then: \[ \begin{aligned} \int\dfrac{2x^2+1}{x^2+2x}\,dx &=\int\left(2+\dfrac{1}{2x}-\dfrac{9}{2(x+2)}\right)\,dx \\ &=2x+\dfrac{1}{2}\ln |x|-\dfrac{9}{2}\ln |x+2|+C \end{aligned} \]

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\(\displaystyle \int \dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)}\,dx\)
See a previous problem.

From a previous problem, \[ \dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)} =\dfrac{-1}{x+2}+\dfrac{2}{(x+2)^2}+\dfrac{x+1}{x^2+1} \]

\(\begin{aligned} \displaystyle\int&\dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)}\,dx \\ &=-\ln |x+2|-\dfrac{2}{x+2}+\dfrac{1}{2}\ln (x^2+1)+\arctan (x)+C \end{aligned}\)

From a previous problem, \[ \dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)} =\dfrac{-1}{x+2}+\dfrac{2}{(x+2)^2}+\dfrac{x+1}{x^2+1} \] Consequently: \[\begin{aligned} \int&\dfrac{5x^2+7x+4}{(x+2)^2(x^2+1)}\,dx =\int\left(\dfrac{-1}{x+2}+\dfrac{2}{(x+2)^2}+\dfrac{x+1}{x^2+1}\right)\,dx \\ &=-\ln |x+2|-\dfrac{2}{x+2}+\dfrac{1}{2}\ln (x^2+1)+\arctan (x)+C \end{aligned}\]

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\(\displaystyle \int \dfrac{2x^3-x^2}{(x^2+1)^2}\,dx\)
See a previous problem.

Break up the integral into \(4\) parts. One integral will need a trig substitution.

\(\displaystyle\int\dfrac{2x^3-x^2}{(x^2+1)^2}\,dx =\ln(x^2+1)-\dfrac{1}{2}\arctan(x)+\dfrac{1}{2}\dfrac{x+2}{x^2+1}+C\)

From a previous problem, \[ \dfrac{2x^3-x^2}{(x^2+1)^2} =\dfrac{2x-1}{x^2+1}+\dfrac{-2x+1}{(x^2+1)^2} \] \[\begin{aligned} \int\dfrac{2x^3-x^2}{(x^2+1)^2}\,dx &=\int\left(\dfrac{2x-1}{x^2+1}+\dfrac{-2x+1}{(x^2+1)^2}\right)\,dx\\ \end{aligned}\] We split this as \(4\) integrals. The first \(3\) are straight forward: \[ \int\dfrac{2x}{x^2+1}\,dx=\ln(x^2+1)+C \] \[ \int\dfrac{-1}{x^2+1}\,dx=-\arctan x+C \] \[ \int\dfrac{-2x}{(x^2+1)^2}\,dx=\dfrac{1}{x^2+1}+C \] The last integral requires the trig substitution \(x=\tan\theta\) and \(dx=\sec^2\theta\,d\theta\): \[\begin{aligned} \int &\dfrac{1}{(x^2+1)^2}\,dx =\int\dfrac{1}{(\tan^2\theta+1)^2}\sec^2\theta\,d\theta \\ &=\int\dfrac{1}{(\sec^2\theta)^2}\sec^2\theta\,d\theta =\int\cos^2\theta\,d\theta \\ &=\int\dfrac{1+\cos(2\theta)}{2}\,d\theta =\dfrac{1}{2}\theta+\dfrac{\sin(2\theta)}{4}+C \\ &=\dfrac{1}{2}\theta+\dfrac{\sin\theta\cos\theta}{2}+C \\ &=\dfrac{1}{2}\arctan x+\dfrac{1}{2}\dfrac{x}{\sqrt{x^2+1}}\dfrac{1}{\sqrt{x^2+1}}+C \\ \\ &=\dfrac{1}{2}\arctan x+\dfrac{1}{2}\dfrac{x}{x^2+1}+C \end{aligned}\] Adding them up, we get: \[\begin{aligned} \int&\dfrac{2x^3-x^2}{(x^2+1)^2}\,dx \\ &=\ln(x^2+1)-\arctan(x)+\dfrac{1}{x^2+1} \\ &\qquad+\dfrac{1}{2}\arctan x+\dfrac{1}{2}\dfrac{x}{x^2+1}+C \\ &=\ln(x^2+1)-\dfrac{1}{2}\arctan(x)+\dfrac{1}{2}\dfrac{x+2}{x^2+1}+C \\ \end{aligned}\]

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\(\displaystyle \int \dfrac{4x^2+7x-3}{(x-2)(x+1)^2}\,dx\)

You need to write out the general expansion, solve for the coefficients and then integrate.

\(\displaystyle \int \dfrac{4x^2+7x-3}{(x-2)(x+1)^2}\,dx =3\ln|x-2|+\ln|x+1|-\dfrac{2}{x+1}+C\)

The general expansion is: \[ \dfrac{4x^2+7x-3}{(x-2)(x+1)^2} = \dfrac{A}{x-2} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2} \] We clear the denominator: \[ 4x^2+7x-3=A(x+1)^2 + B(x-2)(x+1) + C(x-2) \] We solve for the coefficients: \[\begin{aligned} \text{If } x &= 2\text{,}&&\text{ then }\quad 27=9A\text{,}&&\text{ so }A = 3\text{.} \\ \text{If } x &= -1\text{,}&&\text{ then } -6=-3C\text{,}&&\text{ so }C = 2\text{.} \\ \text{If } x &= 0\text{,}&&\text{ then } -3=3 - 2B - 4\text{,}&&\text{ so }B = 1\text{.} \end{aligned}\] Consequently: \[ \dfrac{4x^2+7x-3}{(x-2)(x+1)^2} = \dfrac{3}{x-2} + \dfrac{1}{x+1} + \dfrac{2}{(x+1)^2} \] Finally, we integrate: \[\begin{aligned} \int \dfrac{4x^2+7x-3}{(x-2)(x+1)^2}\,dx &=\int \left(\dfrac{3}{x-2}+\dfrac{1}{x+1}+\dfrac{2}{(x+1)^2}\right)\,dx \\ &=3\ln|x-2|+\ln|x+1|-\dfrac{2}{x+1}+C \end{aligned}\]

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\(\displaystyle \int \dfrac{3x^2+x+6}{x^3+2x^2+4x+8}\,dx\)

Factor the denominator by grouping.

\(\begin{aligned} \int &\dfrac{3x^2+x+6}{x^3+2x^2+4x+8}\,dx \\ &=2\ln|x+2|+\dfrac{1}{2}\ln(x^2+4)-\dfrac{1}{2}\arctan \left(\dfrac{x}{2}\right)+C \end{aligned}\)

The denominator factors by grouping terms: \[ x^3+2x^2+4x+8=x^2(x+2)+4(x+2)=(x+2)(x^2+4) \] So the general expansion is: \[ \dfrac{3x^2+x+6}{x^3+2x^2+4x+8}=\dfrac{A}{x+2}+\dfrac{Bx+C}{x^2+4} \] We clear the denominator: \[ 3x^2+x+6=A(x^2+4)+(Bx+C)(x+2) \] We solve for the coefficients: If \(x=-2\), then \(16=8A\), so \(A=2\). Plugging back gives: \[\begin{aligned} 3x^2+x+6&=2(x^2+4)+(Bx+C)(x+2) \\ x^2+x-2&=(Bx+C)(x+2) \end{aligned}\] If \(x=0\), then \(-2=2C\), so \(C=-1\).
If \(x=1\), then \(0=(B-1)(3)\), so \(B=1\). Consequently: \[ \dfrac{3x^2+x+6}{x^3+2x^2+4x+8}=\dfrac{2}{x+2}+\dfrac{x-1}{x^2+4} \] Finally, we integrate: \[\begin{aligned} \int &\dfrac{3x^2+x+6}{x^3+2x^2+4x+8}\,dx =\int\left(\dfrac{2}{x+2}+\dfrac{x}{x^2+4}+\dfrac{-1}{x^2+4}\right)\\ &=2\ln|x+2|+\dfrac{1}{2}\ln(x^2+4)-\dfrac{1}{2}\arctan \left(\dfrac{x}{2}\right)+C \end{aligned}\]

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\(\displaystyle \int \dfrac{x^3+2x^2}{x^2+2x-3}\,dx\)

Start with long division of polynomials.

\(\displaystyle \int \dfrac{x^3+2x^2}{x^2+2x-3}\,dx =\dfrac{1}{2}x^2+\dfrac{9}{4}\ln|x+3|+\dfrac{3}{4}\ln|x-1|+C\)

By long division: \[ \dfrac{x^3+2x^2}{x^2+2x-3} = x + \dfrac{3x}{x^2+2x-3} \] The denominator factors: \[ x^2+2x-3=(x+3)(x-1) \] We write the general partial fraction expansion the remaining fraction: \[ \dfrac{3x}{x^2+2x-3}=\dfrac{A}{x+3}+\dfrac{B}{x-1} \] We clear the denominator: \[ 3x=A(x-1)+B(x+3) \] We plug in the two roots of the denominator:
If \(x=1\), then \(3=4B\). So \(B=\dfrac{3}{4}\).
If \(x=-3\), then \(-9=-4A\). So \(A=\dfrac{9}{4}\).
Finally, we integrate: \[\begin{aligned} \int \dfrac{x^3+2x^2}{x^2+2x-3} \, dx &=\int \left(x+\dfrac{9}{4(x+3)}+\dfrac{3}{4(x-1)}\right) \, dx \\ &=\dfrac{1}{2}x^2+\dfrac{9}{4}\ln|x+3|+\dfrac{3}{4}\ln|x-1|+C \end{aligned}\]

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\(\displaystyle \int \dfrac{3x+4}{x^2(x^2+1)}\,dx\)

\(\begin{aligned}\displaystyle \int &\dfrac{3x+4}{x^2(x^2+1)}\,dx \\ &=3\ln |x|-\dfrac{4}{x}-\dfrac{3}{2}\ln(x^2+1)-4\arctan x+C \end{aligned}\)

The general expansion is: \[ \dfrac{3x+4}{x^2(x^2+1)} = \dfrac{A}{x} +\dfrac{B}{x^2} + \dfrac{Cx+D}{x^2+1} \] We clear the denominator and expand the right side: \[\begin{aligned} 3x+4&=Ax(x^2+1)+B(x^2+1)+(Cx+D)x^2 \\ &=(A+C)x^3+(B+D)x^2+Ax+B \end{aligned}\] The constant term says \(B=4\).
The linear term says \(A=3\).
The quadratic term says \(B+D=0\). So \(D=-B=-4\).
Finally, the cubic term says \(A+C=0\). So \(C=-A=-3\).
So the integral is: \[\begin{aligned} \int &\dfrac{3x+4}{x^2(x^2+1)}\,dx =\int \left(\dfrac{3}{x} +\dfrac{4}{x^2}-\dfrac{3x}{x^2+1}-\dfrac{4}{x^2+1}\right)\\ &=3\ln |x|-\dfrac{4}{x}-\dfrac{3}{2}\ln(x^2+1)-4\arctan x+C \end{aligned}\]

tj
\(\displaystyle \int_0^1 \dfrac{4x^3+2x}{(x^2+1)^2}\,dx\)

\(\displaystyle \int_0^1 \dfrac{4x^3+2x}{(x^2+1)^2}\,dx=2\ln 2-\dfrac{1}{2}\)

The general expansion is: \[ \dfrac{4x^3+2x}{(x^2+1)^2} = \dfrac{Ax+B}{x^2+1} + \dfrac{Cx+D}{(x^2+1)^2} \] We clear the denominator and expand: \[\begin{aligned} 4x^3+2x &= (Ax+B)(x^2+1)+Cx+D \\ 4x^3+2x &= Ax^3+Bx^2+Ax+Cx+B+D \end{aligned}\] We equate each power of \(x\): \[\begin{aligned} x^3\text{:} \qquad &A=4 \\ x^2\text{:} \qquad &B=0 \\ x\text{:} \qquad &A+C=2 \quad\Rightarrow\quad C=-2 \\ 1\text{:} \qquad &B+D=0 \quad\Rightarrow\quad D=0 \end{aligned}\] So the integral is: \[\begin{aligned} \int_0^1 \dfrac{4x^3+2x}{(x^2+1)^2}\,dx &=\int_0^1 \left( \dfrac{4x}{x^2+1} - \dfrac{2x}{(x^2+1)^2} \right)\\ &=\left[2\ln (x^2+1) + \dfrac{1}{x^2+1}\right]_0^1 \\ &=(2\ln 2 + \dfrac{1}{2}) - (1) = 2\ln 2-\dfrac{1}{2} \end{aligned}\]

tj
Evaluate this improper integral \(\displaystyle \int_0^3 \dfrac{x-2}{x^2-2x-3}\,dx\)

Review your improper integrals.

\(\displaystyle \int_0^3 \dfrac{x-2}{x^2-2x-3}\,dx\quad\) diverges to \(-\infty\).

The denominator factors: \[ x^2-2x-3=(x+1)(x-3) \] So the general expansion is: \[ \dfrac{x-2}{x^2-2x-3}=\dfrac{A}{x+1}+\dfrac{B}{x-3} \] We clear the denominator: \[ x-2 = A(x-3)+B(x+1) \] We solve for the coefficients:
If \(x=3\), then \(1=B(4)\). So \(B=\dfrac{1}{4}\).
If \(x=-1\), then \(-3=A(-4)\). So \(A=\dfrac{3}{4}\).
Finally, we integrate: \[\begin{aligned} \int_0^3 &\dfrac{x-2}{x^2-2x-3} =\int_0^3 \left(\dfrac{3}{4}\dfrac{1}{x+1} + \dfrac{1}{4}\dfrac{1}{x-3}\right)\\ &=\left[\dfrac{3}{4}\ln |x+1|+\dfrac{1}{4}\ln|x-3|\right]_0^3 \\ &=\left(\dfrac{3}{4}\ln |4|+\dfrac{1}{4}\ln|``0^-"|\right) -\left(\dfrac{3}{4}\ln |1|+\dfrac{1}{4}\ln|-3|\right) =-\infty \\ \end{aligned}\] The integral diverges.

tj
If a particle moves along a line with velocity \(v(t)=\dfrac{4t}{(t-4)^2}\) ft/sec where \(t\) is the time in seconds, how far does it travel in the time interval \([0,3]\)

The distance the particle travles is the integral of its velocity: \[ x(3)=\int_0^3 v(t)\, dt \]

\(\displaystyle \int_0^3 \dfrac{4t}{(t-4)^2}dt =12-4\ln4 \text{ ft} \approx6.45 \text{ ft}\)

The distance the particle travles is the integral of its velocity: \[ x(3)=\int_0^3 v(t)\, dt \] To do this integral, we uase a partial fraction expansion. The general expansion is: \[ \dfrac{4t}{(t-4)^2}=\dfrac{A}{t-4}+\dfrac{B}{(t-4)^2} \] We clear the denominator and expand: \[\begin{aligned} 4t=A(t-4)+B \\ 4t=At+(-4A+B) \end{aligned}\] We equate each power of \(t\): \[\begin{aligned} t\text{:}\qquad &A=4\\ 1\text{:}\qquad &-4A+B=0\quad\Rightarrow\quad B=16 \end{aligned}\] Finally, we integrate: \[\begin{aligned} x(3)&=\int_0^3 \dfrac{4t}{(t-4)^2}\,dt =\int_0^3 \dfrac{4}{t-4}+\dfrac{16}{(t-4)^2}\,dt\\ &=\left[4\ln|t-4|-16\dfrac{1}{t-4}\right]_0^3\\ &=4\ln 1-4\ln 4+16-4 \\ &=12-4\ln4 \text{ ft} \approx6.45 \text{ ft} \end{aligned}\]

tj

Review Exercises

Evaluate using any appropriate method.

\(\displaystyle \int \dfrac{x^2-1}{x(x^2+1)}\,dx\)

\(\displaystyle \int \dfrac{x^2-1}{x(x^2+1)}\,dx=-\ln |x|+\ln(x^2+1)+C\)

The general partial fraction expansion is: \[\begin{aligned} \dfrac{x^2-1}{x(x^2+1)} = \dfrac{A}{x} + \dfrac{Bx+C}{x^2+1} \end{aligned}\] We clear the denominator and equate coefficients of each power of \(x\): \[\begin{aligned} x^2-1 = A(x^2+1)+(Bx+C)x = (A+B)x^2+Cx+A \end{aligned}\] Since there is no \(x\) term on the left, \(C=0\).
Because the constant term is \(-1\), \(A=-1\).
Because the \(x^2\) term is just \(x^2\), \(A+B=1\), so \(B=2\). The integral becomes: \[\begin{aligned} \int\dfrac{x^2-1}{x(x^2+1)}\,dx &=\int\left(\dfrac{-1}{x}+\dfrac{2x}{x^2+1}\right) \\ &=-\ln|x|+\ln(x^2+1)+C \end{aligned}\]

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\(\displaystyle \int \dfrac{x^3+3x-1}{x^2+x+2}\,dx\)

Once you've divided, the expression is immediately integrable!

\(\displaystyle \int \dfrac{x^3+3x-1}{x^2+x+2}dx =\dfrac{x^2}{2}-x+\ln(x^2+x+2)+C\)

By long division of polynomials: \[ \dfrac{x^3+3x-1}{x^2+x+2} = x-1+\dfrac{2x+1}{x^2+x+2} \] The last term has a numerator which is just the derivative of the denominator. So: \[\begin{aligned} \int &\dfrac{x^3+3x-1}{x^2+x+2}\,dx =\int \left(x-1+\dfrac{2x+1}{x^2+x+2}\right)\,dx\\ &=\dfrac{x^2}{2}-x+\ln(x^2+x+2)+C \end{aligned}\]

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\(\displaystyle \int_1^{\sqrt{3}} \dfrac{x^2+2x}{x^2(x^2+1)}\,dx\)

First factor the numerator and cancel.

\(\displaystyle \int_1^{\sqrt{3}} \dfrac{x^2+2x}{x^2(x^2+1)}\,dx =\ln \dfrac{3}{2}+\dfrac{\pi}{12}\)

We first cancel an \(x\) from the numerator and the denominator. Then we write the general partial fraction expansion: \[ \dfrac{x^2+2x}{x^2(x^2+1)} =\dfrac{x+2}{x(x^2+1)} =\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1} \] We clear the denominator and collect terms: \[ x+2=A(x^2+1)+Bx^2+Cx =(A+B)x^2+Cx+A \] Because the constant term is \(2\), \(A=2\).
Since the \(x\) term is just \(x\), \(C=1\).
Because there is no \(x^2\) term on the left, \(B=-A=-2\). So the indefinite integral is: \[\begin{aligned} \int \dfrac{x^2+2x}{x^2(x^2+1)}\,dx &=\int \left(\dfrac{2}{x}+\dfrac{-2x+1}{x^2+1}\right)\,dx \\ &=\int \left(\dfrac{2}{x}-\dfrac{2x}{x^2+1}+\dfrac{1}{x^2+1}\right) \\ &=2\ln|x|-\ln(x^2+1)+\arctan(x)+C \end{aligned}\] And the definite integral is: \[\begin{aligned} \int_1^{\sqrt{3}} &\dfrac{x^2+2x}{x^2(x^2+1)}\,dx =\left[\rule{0pt}{10pt}2\ln|x|-\ln(x^2+1)+\arctan(x)\right]_1^{\sqrt{3}} \\ &=(\ln3-\ln(4)+\arctan(\sqrt{3}))-(-\ln(2)+\arctan(1)) \\ &=\ln\dfrac{3}{2}+\dfrac{\pi}{12} \end{aligned}\]

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\(\displaystyle \int \dfrac{2x}{x^2-8x+16}\,dx\)

Partial fractions gives us one way to compute this antiderivative. But notice there is a quicker way. We add zero in the form \(-8+8\) we get: \[\begin{aligned} \int \dfrac{2x}{x^2-8x+16}dx &=\int \dfrac{2x-8+8}{x^2-8x+16}dx \\ &=\int \dfrac{2(x-4)}{(x-4)^2}dx+\int \dfrac{8}{(x-4)^2}dx \end{aligned}\] Each term is now readily integrable. (This trick of adding zero is more helpful than you might imagine!) Work this problem by partial fractions as well.

\(\displaystyle \int \dfrac{2x}{x^2-8x+16}\,dx =2\ln(x-4)-\dfrac{8}{x-4}+C\)

Rather than use a partial fraction decomposition, we recognize that the denominator is the square \(x^2-8x+16=(x-4)^2\) and so add and subtract \(8\) from the numerator. After splitting the integral, one part will have some cancellation. \[\begin{aligned} \int \dfrac{2x}{x^2-8x+16}dx &=\int \dfrac{2x-8+8}{x^2-8x+16}dx \\ &=\int \dfrac{2(x-4)}{(x-4)^2}dx+\int \dfrac{8}{(x-4)^2}dx \\ &=2\ln(x-4)-\dfrac{8}{x-4}+C \end{aligned}\]

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\(\displaystyle \int \dfrac{x^3-5x}{x^4-10x^2+9}\,dx\)

Notice the derivative of the denominator is just a multiple of the numerator.

\(\displaystyle \int \dfrac{x^3-5x}{x^4-10x^2+9}dx =\dfrac{1}{4}\ln|x^4-10x^2+9|+C\)

We make the substitution: \[\begin{aligned} u=x^4-10x^2+9 \qquad du&= 4x^3-20x\,dx \\ &=4(x^3-5x)\,dx \end{aligned}\] Then the integral becomes: \[\begin{aligned} \int \dfrac{x^3-5x}{x^4-10x^2+9}\,dx &=\dfrac{1}{4}\int \dfrac{du}{u} =\dfrac{1}{4}\ln|u| \\ &=\dfrac{1}{4}\ln|x^4-10x^2+9|+C \end{aligned}\]

tj
Consider the rational function \[ f(x)=\dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)} \]
1. Find the general partial fraction expansion for \(f(x)\).
  
  \( \dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)} =\dfrac{A}{x+1}+\dfrac{B(x-2)+C}{(x-2)^2+4} \)
  
  \[ \dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)} =\dfrac{A}{x+1}+\dfrac{B(x-2)+C}{(x-2)^2+4} \]
  
  tj
2. Find the partial fraction expansion for \(f(x)\), i.e. find the coefficients.
  
  This is easiest done using complex numbers.
  
  \[ \dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)} =\dfrac{13}{x+1}+\dfrac{-4(x-2)+48}{(x-2)^2+4} \]
  
  The general partial fraction expansion is \[ \dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)} =\dfrac{A}{x+1}+\dfrac{B(x-2)+C}{(x-2)^2+4} \] We clear the denominator: \[\begin{aligned} 9x^2+160 &=A\left((x-2)^2+4\right) \\ &\quad+(B(x-2)+C)(x+1) \qquad (*) \end{aligned}\] To find \(A\), we plug in the root of the denominator \(x+1\) which is \(x=-1\) into (*): \[ 169 =A\left((-3)^2+4\right)=13A \] So \(A=13\). To find \(B\) and \(C\), we can plug in any four other numbers. However, that's a lot of work. It is easier to plug in one of the complex roots of the denominator \((x-2)^2+4\). The roots are \(x=2\pm2i\). So we plug \(x=2+2i\) into (*): \[ 9(2+2i)^2+160 =(B(2i)+C)(3+2i) \] We multiply both sides by the reciprocal of \(3+2i\) which is \((3+2i)^{-1}=\dfrac{3-2i}{13}\): \[\begin{aligned} B(2i)+C &=(9(2+2i)^2+160)\frac{3-2i}{13} \\ &=(9(4+8i-4)+160)\frac{3-2i}{13} \\ &=(160+72i)\frac{3-2i}{13} \\ &=\frac{1}{13}(624-104i) =48-8i \end{aligned}\] Equating real and imaginary parts: \[ C=48 \quad \text{and} \quad B =-4 \] So the partial fraction expansion is: \[ \dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)} =\dfrac{13}{x+1}+\dfrac{-4(x-2)+48}{(x-2)^2+4} \]
  
  tj
3. Compute \(\displaystyle \int f(x)\,dx\).
  
  Split the quadratic terms in the integral. \[\begin{aligned} \int &\dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)^2}\,dx \\ &=\int \dfrac{13}{x+1}+\dfrac{-4(x-2)}{(x-2)^2+4}+\dfrac{48}{(x-2)^2+4}\,dx \end{aligned}\]
  
  \(\begin{aligned} \int &\dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)^2}\,dx \\ &=13\ln|x+1|-2\ln|(x-2)^2+4|+24\arctan\left(\dfrac{x-2}{2}\right)+C \end{aligned}\)
  
  Using the partial fraction expansion, we compute: \[\begin{aligned} \int &\dfrac{9x^2+160}{(x+1)\left((x-2)^2+4\right)^2}\,dx \\ &=\int \dfrac{13}{x+1}+\dfrac{-4(x-2)}{(x-2)^2+4}+\dfrac{48}{(x-2)^2+4}\,dx \\ &=13\ln|x+1|-2\ln|(x-2)^2+4|+24\arctan\left(\dfrac{x-2}{2}\right)+C \end{aligned}\]
  
  tj