Recall: ā ā ā
ā« u d v = u v ā ā« v d u \displaystyle \int u\,dv=u\,v-\int v\,du ā« u d v = u v ā ā« v d u d u = d u d x d x du=\dfrac{du}{dx}\,dx d u = d x d u ā d x d v = d v d x d x dv=\dfrac{dv}{dx}\,dx d v = d x d v ā d x
b1. Choosing the 'Parts'
b. Log Examples
Compute ā« x 4 ln ā” x d x \displaystyle \int x^4\ln x\,dx ā« x 4 ln x d x
Normally we would take the polynomial x 4 x^4 x 4 u u u ln ā” x \ln x ln x u = ln ā” x d v = x 4 d x d u = 1 x d x v = x 5 5 \begin{array}{ll}
u=\ln x & dv=x^4\,dx \\
du=\dfrac{1}{x}\,dx \quad & v=\dfrac{x^5}{5}
\end{array} u = ln x d u = x 1 ā d x ā d v = x 4 d x v = 5 x 5 ā ā ā« x 4 ln ā” x d x = x 5 5 ln ā” x ā ā« x 5 5 1 x d x = x 5 5 ln ā” x ā ā« x 4 5 d x = x 5 5 ln ā” x ā x 5 2 5 + C \begin{aligned}
\int x^4\ln x\,dx
&=\dfrac{x^5}{5}\ln x-\int \dfrac{x^5}{5}\dfrac{1}{x}\,dx \\
&=\dfrac{x^5}{5}\ln x-\int \dfrac{x^4}{5}\,dx \\
&=\dfrac{x^5}{5}\ln x-\dfrac{x^5}{25}+C
\end{aligned} ā« x 4 ln x d x ā = 5 x 5 ā ln x ā ā« 5 x 5 ā x 1 ā d x = 5 x 5 ā ln x ā ā« 5 x 4 ā d x = 5 x 5 ā ln x ā 2 5 x 5 ā + C ā
We check by differentiating (using the Product Rule):
If ā f ( x ) = x 5 5 ln ā” x ā x 5 2 5 f(x)=\dfrac{x^5}{5}\ln x-\dfrac{x^5}{25} f ( x ) = 5 x 5 ā ln x ā 2 5 x 5 ā f ā² ( x ) = x 4 ln ā” x + x 5 5 1 x ā x 4 5 = x 4 ln ā” x
f'(x)=x^4\ln x+\dfrac{x^5}{5}\dfrac{1}{x}-\dfrac{x^4}{5}
=x^4\ln x
f ā² ( x ) = x 4 ln x + 5 x 5 ā x 1 ā ā 5 x 4 ā = x 4 ln x
[Ć]
Compute the integral ā« x 9 ln ā” x d x \displaystyle \int x^9\ln x\,dx ā« x 9 ln x d x
Answer
ā« x 9 ln ā” x d x = 1 1 0 x 1 0 ln ā” x ā 1 1 0 0 x 1 0 + C \displaystyle \int x^9\ln x\,dx
=\dfrac{1}{10}x^{10}\ln x-\dfrac{1}{100}x^{10}+C ā« x 9 ln x d x = 1 0 1 ā x 1 0 ln x ā 1 0 0 1 ā x 1 0 + C
[Ć] Solution
We let:
u = ln ā” x d v = x 9 d x d u = 1 x d x v = 1 1 0 x 1 0 \begin{array}{ll}
u=\ln x & dv=x^9\,dx\\
du=\dfrac{1}{x}\,dx \quad & v=\dfrac{1}{10}x^{10}
\end{array} u = ln x d u = x 1 ā d x ā d v = x 9 d x v = 1 0 1 ā x 1 0 ā ā« x 9 ln ā” x d x = 1 1 0 x 1 0 ln ā” x ā ā« 1 1 0 x 9 d x = 1 1 0 x 1 0 ln ā” x ā 1 1 0 0 x 1 0 + C \begin{aligned}
\int x^9\ln x\,dx
&=\dfrac{1}{10}x^{10}\ln x-\int \dfrac{1}{10}x^9\,dx \\
&=\dfrac{1}{10}x^{10}\ln x-\dfrac{1}{100}x^{10}+C
\end{aligned} ā« x 9 ln x d x ā = 1 0 1 ā x 1 0 ln x ā ā« 1 0 1 ā x 9 d x = 1 0 1 ā x 1 0 ln x ā 1 0 0 1 ā x 1 0 + C ā
[Ć] Check
We check using the Product Rule:
d d x ( 1 1 0 x 1 0 ln ā” x ā 1 1 0 0 x 1 0 ) = x 9 ln ā” x + 1 1 0 x 1 0 1 x ā 1 1 0 x 9 = x 9 ln ā” x \begin{aligned}
\dfrac{d}{dx}&\left(\dfrac{1}{10}x^{10}\ln x-\dfrac{1}{100}x^{10}\right) \\
&=x^9\ln x+\dfrac{1}{10}x^{10}\dfrac{1}{x}-\dfrac{1}{10}x^9
=x^9\ln x
\end{aligned} d x d ā ā ( 1 0 1 ā x 1 0 ln x ā 1 0 0 1 ā x 1 0 ) = x 9 ln x + 1 0 1 ā x 1 0 x 1 ā ā 1 0 1 ā x 9 = x 9 ln x ā
[Ć]
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