Anti-derivative is consider as the reverse process of differentiation, which, for a given derivative essentially recovers the original function. Anti-derivative has many uses. For example, it is used by a physicist who knows the velocity of a particle to know its position at a given time. And an engineer who can measure the variable rate at which the water is leaking from a tank wants to know the amount leaked over certain time period. In each case, the problem is to find a function $F$ whose derivative is a known function $f$. If such a function $F$ exists, it is called an anti-derivative of $f$. In this section we will be studying more about anti-derivatives.

A function $F$ is an anti-derivative of a function $f$ when for every $x$ in the domain of $f$, it follows that $F'(x) = f(x)$.

For example, since the derivative of $x^{3}$ is $3x$, an anti-derivative of $3x$ is $x^{3}$. There are, however other anti-derivative of $3x$. Each of the following is an anti-derivative of $3x$:

$x^{3}+1$, $x^{3}-20$, $x^{3}+e$

Clearly, we may add any constant to $x^{3}$ and the derivative will be $3x$. Therefore,$x^{3}+C$ is the anti-derivative of $3x$ for any constant C.

$\int 3x dx$ = $x^{3}+C$
Here the function to be integrated (here $3x$) is called the integrand. The variable reminds us that the variable of integration is $x$. The constant $C$ is called the arbitrary constant because it may take any value positive, negative or zero.

$\int x^{n}dx$ = $\frac{x^{n+1}}{n+1}$ + C (n ? -1)

$\int \frac{1}{x}dx$ = $ln |x| + C ( x \neq 0)$

$\int e^{x}dx$ = $e^{x} + C$

$\int e^{-x}dx$ = $-e^{-x} + C$

$\int a^{x}dx$ = $\frac{a^{x}}{ln(a)}$ + C

$\int sinx dx$ = $-cos x + C$

$\int cosx dx$ = $sin x + C$

$\int sec^{2}xdx$ = $tan x + C$

$\int csc^{2}xdx$ = $-cot x + C$

$\int sec(x)tan(x)dx$ = $sec x + C$

$\int csc(x)cot(x)dx$ = $-csc x + C$

$\int \frac{1}{\sqrt{1-x^{2}}}dx$ = $sin^{-1} x + C$

$\int \frac{1}{1+x^{2}}dx$ = $tan^{-1} x + C$

$\int \frac{1}{|x|\sqrt{x^{2}-1}}$ = $sec^{-1} x + C$

Solved Examples

Question 1: Find the anti-derivative for the function $f(x)$ = $\frac{1}{2}$
Given function $f(x)$ = $\frac{1}{2}$

 $\int \frac{1}{2}$$dx$ = $\frac{1}{2}$$\int dx$

= $\frac{1}{2}$$x^{1}$

= $\frac{x}{2}$


Question 2: Find the anti-derivative for the function $f(x)$ = $3x$
Given function $f(x)$ = $3x$

$\int 3x dx$ = $3$ $\int x dx$

= $3$ $\int x^{1} dx$

= $3$ $(\frac{x^{2}}{2})$$+C$

= $\frac{3}{2}$ $x^{2}+C$


Question 3: Find the anti-derivative for the function $f(x)$ = $\frac{2}{x^{7}}$
Given function $f(x)$ = $\frac{2}{x^{7}}$

$\int f(x) dx$ = $\int \frac{2}{x^{7}}$$dx$

= 2 $\int \frac{1}{x^{7}}$$dx$

= $2 \int x^{-7}d$

= $2$ $\frac{x^{-7+1}}{-7+1}$

=  $2$ $\frac{x^{-6}}{-6}$

=  $-\frac{1}{3x^{6}}$$+C$

Question 4: Find the anti-derivative of trigonometric function $f(x)$ = $8cos x  + 2tan2x$  with respect to x
Given $f(x)$ = $8cos x  + 2tan2x$
$f(x)$ = $\int (8cos x + 2tan2x) dx$  = $\int 8cos x dx$ + $\int 2tan2x dx$
                                                                     = $8 \int(cos x dx) + 2 \int (tan2x) dx$
                                                                     = $8 \int (cos x dx) + 2 (sec2x - 1) dx$
                                                                     = $8 \int (cos x dx) + 2 \int (sec2x) dx -  2\int dx$
                                                                     = $8 (sin x) + 2 tan x - 2x+ c$
                                                                     = $8 sin x + 2 tan x  - 2x + c$
   $\int(8cos x  + 3tan2x) dx$ =  $8sin x + 2 tan x - 2x+ c$