Question

How do you integrate?

Answer 1

Use the substitution `x+1=sqrt2tantheta`.

Explanation:

Let

`I=intx^2/sqrt(x^2+2x+3)dx`

Complete the square in the square root:

`I=intx^2/sqrt((x+1)^2+2)dx`

Apply the substitution `x+1=sqrt2tantheta`:

`I=int(sqrt2tantheta-1)^2/(sqrt2sectheta)(sqrt2sec^2thetad theta)`

Simplify:

`I=int(2tan^2theta-2sqrt2tantheta+1)secthetad theta`

Rearrange:

`I=int(2sec^3theta-2sqrt2secthetatantheta+3sectheta)d theta`

These are all known integrals:

`I=(secthetatantheta-ln|sectheta+tantheta|)-2sqrt2sectheta+3ln|sectheta+tantheta|+C`

Rearrange:

`I=1/2(sqrt2tantheta-4)(sqrt2sectheta)+2ln|sqrt2tantheta+sqrt2sectheta|+C`

Reverse the substitution:

`I=1/2(x-3)sqrt(x^2+2x+3)+2ln|(x+1)+sqrt(x^2+2x+3)|+C`