Question

How do you find the integral `int sqrt(x^3+x^2)(3x^2+2x)dx` using substitution?

Answer 1

Let `u = x^3 + x^2`. Then `du = 3x^2 +2xdx` and `dx= (du)/(3x^2 + 2x)`.

`I = int sqrt(u)(3x^2 + 2x)/(3x^2 + 2x) du`

`I = int sqrt(u) du`

`I = 2/3u^(3/2) + C`

`I = 2/3(x^3 + x^2)^(3/2) + C`

Hopefully this helps!

Answer 2

`I=intsqrt(x^3+x^2)(3x^2+2x)dx` `=int(x^3+x^2)^(1/2)d/(dx)(x^3+x^2)dx=(x^3+x^2)^(1/2+1)/(1/2+1) +c`
`=2/3(x^3+x^2)^(3/2)+c`

Explanation:

Here,

`I=intsqrt(x^3+x^2)(3x^2+2x)dx`

Substituting

`x^3+x^2=u^2=>(3x^2+2x)dx=2udu`

`=>I=intsqrt(u^2)2udu`

`=intu2udu`

`=2intu^2 du`

`=2[u^3/3 ]+c`

`=2/3(u^2)^(3/2)+c`

`=2/3(x^3+x^2)^(3/2)+c`