Question

How do you use substitution to integrate `x^2sqrt(x+1)`?

Answer 1

Refer to explanation

Explanation:

We set `u=sqrt(x+1)` hence `du=1/(2*sqrt(x+1))dx`

#int x^2*sqrt(x+1)dx=2*int (u^2-1)^2*u^2du= 2*int (u^6-2u^4+u^2)du=2*[u^7/7-2*u^5/5+u^3/3] #

Now replacing u with `sqrt(x+1)` we get finally that

`int x^2*sqrt(x+1)dx=2/105*sqrt((x+1)^3)*(15x^2-12x+8)+c`