Question

How do you integrate `int xsqrt(3x+ 1) dx`?

Answer 1

`int xsqrt(3x + 1) = 2/9x(3x + 1)^(3/2) - 4/135(3x + 1)^(5/2) + C`

Explanation:

You will need substitution in addition to integration by parts. Let `u = x` and `dv = sqrt(3x +1)`. By the power rule, `du = dx`. But we will need to make a substitution to find `v`. Let `t= 3x+ 1`. Then `dt = 3dx` and `dx = (dt)/3`.

`intsqrt(3x + 1) = sqrt(t) * (dt)/3 = 1/3sqrt(t)dt = 2/9t^(3/2) = 2/9(3x + 1)^(3/2)`

Apply integration by parts now.

`int udv = uv - int vdu`

`intxsqrt(3x+ 1) = 2/9x(3x + 1)^(3/2) - int 2/9(3x+ 1)^(3/2)dx`

`int xsqrt(3x + 1) = 2/9x(3x + 1)^(3/2) - 2/9 int (3x + 1)^(3/2)dx`

Now let `n = 3x + 1`. Then `dn =3dx` and `dx = (dn)/3`.

`int (3x + 1)^(3/2) = 1/3int n^(3/2) dn = 2/15n^(5/2) = 2/15(3x + 1)^(5/2)`

`int xsqrt(3x + 1) = 2/9x(3x + 1)^(3/2) - 2/9(2/15)(3x +1)^(5/2) + C`

`int xsqrt(3x + 1) = 2/9x(3x + 1)^(3/2) - 4/135(3x + 1)^(5/2) + C`

Hopefully this helps!