# How do you evaluate the integral int(1+x)^2 dx?

Aug 15, 2014

$= {x}^{3} / 3 + {x}^{2} + x + C$, where $C$ is a constant

Explanation :

$I = \int {\left(1 + x\right)}^{2} \mathrm{dx}$

It can be solved by two methods,

$\left(I\right)$

let's $1 + x = t$ $\implies$ $\mathrm{dx} = \mathrm{dt}$

then, $\int {t}^{2} \mathrm{dt} = {t}^{3} / 3 + c$

Substituting $t$ back,

$= {\left(1 + x\right)}^{3} / 3 + c$, where $c$ is a constant

$= \frac{1}{3} \left({x}^{3} + 3 {x}^{2} + 3 x + 1\right) + c$, where $c$ is a constant

$= {x}^{3} / 3 + {x}^{2} + x + \frac{1}{3} + c$, where $c$ is a constant

$= {x}^{3} / 3 + {x}^{2} + x + C$, where $C$ is again a constant

$\left(I I\right)$

Expanding ${\left(1 + x\right)}^{2} = {x}^{2} + 2 x + 1$, we get

$\int \left({x}^{2} + 2 x + 1\right) \mathrm{dx}$

$= {x}^{3} / 3 + 2 {x}^{2} / 2 + x + c$, where $c$ is a constant

$= {x}^{3} / 3 + {x}^{2} + x + c$, where $c$ is a constant