Question

How do you integrate `int xsqrt(2x+1)` using substitution?

Answer 1

`= 1/10(2x+1)^(5/2) - 1/6(2x+1)^(3/2) + C`

Explanation:

Let `u = 2x+1 implies du = 2dx`

`implies x = (u-1)/2 and dx = (du)/2`

`int xsqrt(2x+1)dx = 1/4 int (u-1)u^(1/2)du`

`= 1/4int u^(3/2) - u^(1/2)du`

`=1/4[2/5u^(5/2) - 2/3u^(3/2)] + C`

`= 1/10(2x+1)^(5/2) - 1/6(2x+1)^(3/2) + C`