21. Antiderivatives, Areas and the FTC

b. Antiderivative Rules

2. Exponential and Logarithmic Functions

The table on the previous page listed the properties of antiderivatives with respect to algebraic operations. The table on this page gives the antiderivatives for exponential and logarithmic functions.

Derivatives

We first review their derivatives:

Exponential Functions	Logarithmic Functions
\(\dfrac{d}{dx}e^x=e^x\)	\(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\)
\(\dfrac{d}{dx}b^x=\ln(b)\,b^x\)	\(\dfrac{d}{dx}\log_b x=\dfrac{1}{\ln b}\,\dfrac{1}{x}\)

Antiderivatives

Here are their antiderivatives. You can check each line of the table by differentiating the quantity on the right to see you get the quantity on the left.

	If the Function is	then the Antiderivative is
Natural Exponential	\(f(x)=e^x\)	\(F(x)=e^x+C\)
General Exponential	\(f(x)=b^x\)	\(F(x)=\dfrac{b^x}{\ln b}+C\)
Natural Logarithm\(^\text{1}\)	\(f(x)=\dfrac{1}{x}\)	\(F(x)=\ln\|x\|+C\)
General Logarithm\(^\text{2}\)	\(f(x)=\dfrac{1}{x}\)	\(F(x)=(\ln b)\log_b \|x\|+C\)

\(^\text{1}\) This is the function whose antiderivative is \(\ln x\). We will discuss the antiderivatives of \(\ln x\) and \(\log_b x\) in Calculus 2 in the chapter on Integration by Parts.
\(^\text{2}\) This antiderivative formula is not useful, since the formula using \(\ln x\) is equivalent and easier.

Why is there an absolute value in the antiderivatives which have logs?

Like the function \(\sqrt{x}\), the function \(\ln{x}\) is only defined for positive values of \(x\), since it is the inverse function of \(e^x\) which is always positive. However, the function \(\dfrac{1}{x}\) is defined for all \(x\ne0\) and we might want to know its antiderivative even when \(x\) is negative.

For \(x \gt 0\) we have \(\ln x=\ln|x|\) and so \[ \dfrac{d}{dx}\ln|x|=\dfrac{d}{dx}\ln x=\dfrac{1}{x} \] Thus, \(\ln|x|\) is an antiderivative of \(\dfrac{1}{x}\) without or with the absolute value.

However, for \(x \lt 0\), we need to find an antiderivative. We work backwards: For \(x \lt 0\), we have \(|x|=-x\) and so \(\ln|x|=\ln(-x)\). We differentiate using the Chain Rule: \[\begin{aligned} \dfrac{d}{dx}\ln|x|&=\dfrac{d}{dx}\ln(-x)=\dfrac{1}{(-x)}\dfrac{d}{dx}(-x) \\ &=\dfrac{1}{(-x)}(-1)=\dfrac{1}{x} \end{aligned}\] Reversing this, we have \(\ln|x|\) is an antiderivative of \(\dfrac{1}{x}\) but we have to use the absolute value!

We conclude \(\ln|x|\) is an antiderivative of \(\dfrac{1}{x}\) for both positive or negative values of \(x\).

Find the general antiderivative of \(g(t)=3t^2e^{\small t^3}\).

\(G(t)=e^{\small t^3}+C\).

Since \(\dfrac{d}{dt}t^3=3t^2\), the Chain Rule says \(g(t)\) is the derivative of \(G(t)=e^{\small t^3}+C\).