Evaluate the following integral `x^3sqrt(1+x^2)`?

`int x^3 (1+x^2)^(1/2) dx`

Let `u = 1+x^2`.

This make `du = 2x dx` and `x^2 = u-1`

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

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

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

`= 1/15 [3(1+x^2)^(5/2) - 5(1+x^2)^(3/2)] +C`

`= 1/15 (1+x^2)^(3/2)[3+3x^2-5 ] +C`

`= 1/15 (x^2-1)^(3/2) ( 3x^2-2) +C`

`int x^3 (1+x^2)^(1/2) dx`

We can integrate `t = (1+x^2)^(1/2)` by substitution if we think of one of the `x`'s as attached to `dx` to be `dt = 2x dx`.

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

Let `u = x^2` and `dv = (1+x^2)^(1/2) x dx`,

so `du = 2x dx` and `v = 1/3(1+x^2)^(3/2)`

`uv-int v du = 1/3 x^2 (1+x^2)^(3/2) - 2/3 int x (1+x^2)^(3/2) dx`

We can evaluate the integral by substitution, to get

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

`= 1/15(x^2 +1)^(3/2)[5x^2-2(x^2+1)]`

`= 1/15(x^2 +1)^(3/2)(3x^2-2)`

Let me try another approach:

`d/(dx) x^n (1+x^2)^(3/2) = nx^(n-1) (1+x^2)^(3/2) + 3/2x^n(2x) (1+x^2)^(1/2)`

`color(white)(d/(dx) x^n (1+x^2)^(3/2)) = nx^(n-1) (1+x^2)^(3/2) + 3x^(n+1) (1+x^2)^(1/2)`

`color(white)(d/(dx) x^n (1+x^2)^(3/2)) = (nx^(n-1) (1+x^2)+3x^(n+1))(1+x^2)^(1/2)`

`color(white)(d/(dx) x^n (1+x^2)^(3/2)) = ((n+3)x^(n+1)+nx^(n-1))(1+x^2)^(1/2)`

In particular, putting `n=2` then `n=0` we find:

`d/(dx) (1/5x^2(1+x^2)^(3/2)) = (x^3+2/5x)(1+x^2)^(1/2)`

`d/(dx) (-2/15(1+x^2)^(3/2)) = -2/5x(1+x^2)^(1/2)`

Hence:

`int x^3(1+x^2)^(1/2) dx = (1/5x^2-2/15)(1+x^2)^(3/2) + C`

`color(white)(int x^3(1+x^2)^(1/2) dx) = 1/15(3x^2-2)(1+x^2)^(3/2) + C`