# How to Interpret and Calculate the Indefinite Integral

Integrals are mainly split into two categories: Definite and indefinite integrals. The indefinite integral is the same as the anti-derivative.

A challenge in working with integrals is having to find the expression without determining the constant term. The symbol $C$ that you add to the end of integrals represents this unknown constant of integration. To find the constant, you are dependent on having constraints, such that you can find $C$ by substitution and equation solving.

Theory

### TheIndefiniteIntegral

 $\int f\left(x\right)\phantom{\rule{0.17em}{0ex}}dx=F\left(x\right)+C,\phantom{\rule{2em}{0ex}}{F}^{\prime }\left(x\right)=f\left(x\right)$

Here, $f\left(x\right)$ is called the integrand and $C$ is called the constant of integration.

Example 1

Compute the integral $\int \mathrm{ln}x+{e}^{x}+{x}^{3}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill \int \mathrm{ln}x+{e}^{x}+{x}^{3}\phantom{\rule{0.17em}{0ex}}dx& =x\mathrm{ln}x-x+{e}^{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\frac{1}{4}{x}^{4}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill \int \mathrm{ln}x+{e}^{x}+{x}^{3}\phantom{\rule{0.17em}{0ex}}dx=x\mathrm{ln}x-x+{e}^{x}+\frac{1}{4}{x}^{4}+C& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Example 2

Compute the integral

 $\int 3\mathrm{cos}\left(3x\right)-4\mathrm{sin}\left(2x\right)\phantom{\rule{0.17em}{0ex}}dx$

$\int 3\mathrm{cos}\left(3x\right)-4\mathrm{sin}\left(2x\right)\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill & \phantom{=}\int 3\mathrm{cos}\left(3x\right)-4\mathrm{sin}\left(2x\right)\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot \frac{1}{3}\mathrm{sin}\left(3x\right)+4\cdot \frac{1}{2}\mathrm{cos}\left(2x\right)+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\left(3x\right)+2\mathrm{cos}\left(2x\right)+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \int 3\mathrm{cos}\left(3x\right)-4\mathrm{sin}\left(2x\right)\phantom{\rule{0.17em}{0ex}}dx& =3\cdot \frac{1}{3}\mathrm{sin}\left(3x\right)+4\cdot \frac{1}{2}\mathrm{cos}\left(2x\right)+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\left(3x\right)+2\mathrm{cos}\left(2x\right)+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Compute the integral

 $\int \mathrm{sin}\left(2x\right)+3\mathrm{cos}\left(x\right)-{e}^{5x}\phantom{\rule{0.17em}{0ex}}dx$

$\int \mathrm{sin}\left(2x\right)+3\mathrm{cos}\left(x\right)-{e}^{5x}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill & \phantom{=}\int \mathrm{sin}\left(2x\right)+3\mathrm{cos}\left(x\right)-{e}^{5x}\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-\frac{1}{2}\mathrm{cos}\left(2x\right)+3\mathrm{sin}\left(x\right)-\frac{1}{5}{e}^{5x}+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\int \mathrm{sin}\left(2x\right)+3\mathrm{cos}\left(x\right)-{e}^{5x}\phantom{\rule{0.17em}{0ex}}dx=-\frac{1}{2}\mathrm{cos}\left(2x\right)+3\mathrm{sin}\left(x\right)-\frac{1}{5}{e}^{5x}+C$