Finding antiderivatives

Key to solving integrals, as we have seen, is to find antiderivatives. Here we discuss some methods of how to find them for the basic functions. In general, however, it is not trivial at all to find for any given function an antiderivative! It helps to think about finding the antiderivative as going up, while finding the derivative as going down :

\begin{array}{lll} F & & F\\ \scriptstyle\prime\big\downarrow & & \big\uparrow\scriptstyle\int\\ f & & f \\ \end{array}

Note the integral sign $\int$ to the right of the arrow. It is often used to indicate the process of finding the antiderivative. The notion of going up for the antiderivative is enhanced by using capital letters for antiderivatives (but this is not a rule).

So how do we find the antiderivative of a specific function $f$ ? The general strategy is to go through every function you know and check if its derivative is the desired function $f$ :

\begin{array}{lll} ? \\ \scriptstyle\prime\big\downarrow \\ f \\ \end{array}

In this way we quickly find the antiderivative of some basic functions:

Theorem 1

Equation 1

\begin{array}{ccccccc} e^x & \sin(x) & -\cos(x) & \ln(x)\\ {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int} & {\tiny\prime}\!{\big\downarrow}\,{\big\uparrow}\!{\tiny\int} & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int} & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int}\\ e^x & \cos(x) & \sin(x) & x^{-1} \\ \end{array}

Antiverivatives of exponential, logarithmic and trigonometric functions.

And we also see that finding the antiderivative of weighted sum is no problem:

Theorem 2

The antiderivative of a weighted sum $ag(x)+bh(x)$ is the weighted sum of the antiderivatives:

Equation 2

\begin{array}{cccc} G & H & & a\cdot G+ b\cdot H\\ {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int} & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int} & \Rightarrow & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int}\\ g & h & & a\cdot g+ b\cdot h \\ \end{array}

Antiderivative of a weight sum is the sum of the antiderivatives.

Example 1

Find the antiderivative of the function $f(x)=3\cos(x)-5e^x$ .

Solution

\begin{array}{cccc} \sin(x) & e^x & & 3\sin(x)-5 e^x\\ {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int} & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int} & \Rightarrow & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int}\\ \cos(x) & e^x & & 3\cos(x)-5 e^x\\ \end{array}

Finally, let's discuss the antiderivative of power functions $x^n$ . We know that the derivative of a power function is essentially obtained by lowering the power by one, so a candidate for the antiderivative is $x^{n+1}$ , but have to apply a multiplication factor:

Theorem 3

To find the antiderivative of $x^n$ , increase first the exponent by $1$ , and then divide by the new exponent.

Equation 3

\begin{array}{ccccccc} & & \frac{1}{n+1} x^{n+1} & & & & \ln(x) \\[0.3em] n\neq -1 & \Rightarrow & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int} & & n=-1 & \Rightarrow & {\tiny\prime}\!{\big\downarrow}\!\!{\big\uparrow}\!{\tiny\int}\\ & & x^n & & & & x^{-1} \\ \end{array}

Antiderivative of a power function.

Note that this rule does not for $n=-1$ , as increasing the exponent by $1$ results in $0$ as the new exponent, and we cannot divide by $0$ . But we already now from differential calculus, that the derivative of $\ln(x)$ is $\frac{1}{x}=x^{-1}$ .

Exercise 1

Determine the antiderivative of the following functions:

$f(x)=3$ for all $x$ (constant function)
$f(x)=2x$
$f(x)=\sqrt{x}$
$f(x)=\frac{1}{x^2}$
$f(x)=\frac{2}{x}$
$f(x)=2x^3+4x+1$
$f(x)=(x-1)^2$
$f(x)=\frac{2}{x}+3e^x$

Solution

$f(x)=3x^0$ (thus $n=0$ ). It follows $F(x)=3\cdot \frac{1}{0+1} x^{0+1}=3x$
$f(x)=2 x^1$ (thus $n=1$ ). It follows $F(x)=2\cdot \frac{1}{1+1}x^{1+1}=x^2$
$f(x)=x^{0.5}$ (thus $n=0.5$ ). It follows $F(x)=\frac{1}{1+0.5}x^{1+0.5} = \frac{2}{3}x^{3/2}$
$f(x)=x^{-2}$ (thus $n=-2$ ). It follows $F(x)=\frac{1}{-2+1}x^{-2+1} =-x^{-1}=-\frac{1}{x}$ .
$f(x)=2 x^{-1}$ (thus $n=-1$ ). It follows $F(x)=2\ln(x)$
$F(x)=2\cdot\frac{1}{3+1}x^{3+1}+4\cdot \frac{1}{1+1} x^{1+1} + 1\cdot\frac{1}{0+1}x^{0+1}=\frac{1}{2}x^4+2x^2+x$
$f(x)=x^2-2x+1$ , thus $F(x)=\frac{1}{3}x^3-2\cdot \frac{1}{2}x^2+\frac{1}{1}x=\frac{1}{3}x^3-x^2+x$
$f(x)=2\frac{1}{x}+3e^x$ , thus $F(x)=2\ln(x)+3e^x$