Integrals of exponential and logarithmic functions

Exponential and logarithmic functions, grade 12 (5 units) Lesson 20 of 23Learning

Integrals of exponential and logarithmic functions

Recap: integrals from grade 11

In grade 11 you learnt:

$\displaystyle F$ is an antiderivative of $\displaystyle f$ when $\displaystyle {F'(x)}={f(x)}$ ; all the antiderivatives are written as the indefinite integral, with $\displaystyle +C$ .
The integral of a sum is the sum of the integrals, and a number that multiplies a function comes out of the integral.
When the inner function is linear, you divide by the coefficient of $\displaystyle x$ :

\displaystyle \int f(mx+b)\,dx=\frac{1}{m}F(mx+b)+C

The definite integral: $\displaystyle {\int_a^b f(x)\,dx}={F(b)-F(a)}$ .

Every new derivative from the previous lessons gives a new integral: you reverse the rule of differentiation.

The integrals of

\displaystyle e^x

and of

\displaystyle e^{mx+b}

$\displaystyle {(e^x)'}={e^x}$ , so $\displaystyle e^x$ is an antiderivative of itself:

\displaystyle \int e^{x}\,dx=e^{x}+C

\displaystyle \int e^{mx+b}\,dx=\frac{e^{mx+b}}{m}+C

The example below finds the integral of a sum, term by term:

\int\left(4e^{2x}-e^{-x}\right)dx=4\cdot\frac{e^{2x}}{2}-\frac{e^{-x}}{-1}+C=2e^{2x}+e^{-x}+C

A check by differentiating: $\displaystyle {\left(2e^{2x}+e^{-x}\right)'}={4e^{2x}-e^{-x}}$ .

The integral of

\displaystyle a^x

$\displaystyle {(a^x)'}={a^x\cdot\ln a}$ , so the derivative of $\displaystyle \frac{a^x}{\ln a}$ is $\displaystyle a^x$ :

\displaystyle \int a^{x}\,dx=\frac{a^{x}}{\ln a}+C

$\displaystyle {\int3^x\,dx}={\frac{3^x}{\ln3}+C}$ .
$\displaystyle {\int_0^1 2^x\,dx}={\left[\frac{2^x}{\ln2}\right]_0^1}={\frac{2-1}{\ln2}}={\frac{1}{\ln2}\approx1.44}$ .

The integrals of

\displaystyle \frac{1}{x}

and of

\displaystyle \frac{1}{mx+b}

$\displaystyle {(\ln x)'}={\frac{1}{x}}$ where $\displaystyle x>0$ , and also $\displaystyle {\left(\ln(-x)\right)'}={\frac{-1}{-x}}={\frac{1}{x}}$ where $\displaystyle x<0$ . The two cases together are written with an absolute value:

\displaystyle \int\frac{1}{x}\,dx=\ln|x|+C

\displaystyle \int\frac{1}{mx+b}\,dx=\frac{\ln|mx+b|}{m}+C

$\displaystyle {\int\frac{6}{2x-1}\,dx}={6\cdot\frac{\ln|2x-1|}{2}+C}={3\ln|2x-1|+C}$ .
Where $\displaystyle x>0$ you may write $\displaystyle \ln x$ without the absolute value.

An integral by the inner derivative

When the integral holds an inner function $\displaystyle u(x)$ and also its derivative $\displaystyle u'(x)$ , the antiderivative is a composite function:

\displaystyle \int f(u(x))\cdot u'(x)\,dx=F(u(x))+C

Two important cases:

\displaystyle \int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C

\displaystyle \int f'(x)\cdot e^{f(x)}\,dx=e^{f(x)}+C

In the examples below we spot the inner function and its derivative:

$\displaystyle \int\frac{2x}{x^2+1}\,dx$ : the derivative of $\displaystyle x^2+1$ is $\displaystyle 2x$ , so the result is $\displaystyle \ln(x^2+1)+C$ . $\displaystyle x^2+1>0$ , so no absolute value is needed.
$\displaystyle \int xe^{x^2}\,dx$ : the derivative of $\displaystyle x^2$ is $\displaystyle 2x$ , and the integral holds only $\displaystyle x$ – half of it. So $\displaystyle {\int xe^{x^2}\,dx}={\frac{1}{2}\int2xe^{x^2}\,dx}={\frac{1}{2}e^{x^2}+C}$ .

A check by differentiating: $\displaystyle {\left(\frac{1}{2}e^{x^2}\right)'}={\frac{1}{2}\cdot2x\cdot e^{x^2}}={xe^{x^2}}$ .

The integral of a rational function

When the numerator’s degree is at least the denominator’s, split the fraction into a sum – each term is a known integral. In the examples below:

$\displaystyle {\int\frac{x^2+1}{x}\,dx}={\int\left(x+\frac{1}{x}\right)dx}={\frac{x^2}{2}+\ln|x|+C}$ .
$\displaystyle {\frac{2x+3}{x+1}}={\frac{2(x+1)+1}{x+1}}={2+\frac{1}{x+1}}$ , so $\displaystyle {\int\frac{2x+3}{x+1}\,dx}={2x+\ln|x+1|+C}$ .

A function from its derivative and a point

Of all the antiderivatives, only one passes through a given point. The example below finds the function $\displaystyle f$ whose derivative is $\displaystyle {f'(x)}={e^{2x}+\frac{1}{x}}$ for $\displaystyle x>0$ , and whose graph passes through the point $\displaystyle \left(1,\frac{e^2}{2}+2\right)$ :

$\displaystyle {f(x)}={\frac{e^{2x}}{2}+\ln x+C}$ .
We substitute the point: $\displaystyle {f(1)}={\frac{e^2}{2}+\ln1+C}={\frac{e^2}{2}+C}$ , so $\displaystyle {\frac{e^2}{2}+C}={\frac{e^2}{2}+2}$ and $\displaystyle {C}={2}$ .
$\displaystyle {f(x)}={\frac{e^{2x}}{2}+\ln x+2}$ .

Exercise 1

Find the integrals.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power