Question

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

Answer 1

`=x^3/3+x^2+x+C`, where `C` is a constant

Explanation :

`I=int(1+x)^2dx`

It can be solved by two methods,

`(I)`

using Integration by Substitution

let's `1+x=t` `=>` `dx=dt`

then, `intt^2dt=t^3/3+c`

Substituting `t` back,

`=(1+x)^3/3+c`, where `c` is a constant

`=1/3(x^3+3x^2+3x+1)+c`, where `c` is a constant

`=x^3/3+x^2+x+1/3+c`, where `c` is a constant

`=x^3/3+x^2+x+C`, where `C` is again a constant

`(II)`

Expanding `(1+x)^2=x^2+2x+1`, we get

`int(x^2+2x+1)dx`

`=x^3/3+2x^2/2+x+c`, where `c` is a constant

`=x^3/3+x^2+x+c`, where `c` is a constant