Integration with inverse trigonometric substitution

[Toutatis]

Homework Statement

The problem is: integrate: (1/(4x²+4x+5)²)dx

Homework Equations

The Attempt at a Solution

4x²+4x+5=(2x+1)²+4 gives:

∫dx/((2x+1)²+4)²

use regular substitution:
u=2x+1
du=2dx
dx=1/2du
gives: 1/2∫du/(u²+4)²)

trigonometric substitution:
u=tan(z)
du=1/(cos(z))²dz
gives: 1/2∫dz/((sin²(z)/cos²(z)) +4)²

I have tried most trigonometric tricks I can think off to make that last integral into something that I know how to integrate, but it has usually just gotten much worse, and certainly not better. I think it will be helpful to show my different attempts beyond this point.

 

[tiny-tim]

… gives: 1/2∫du/(u²+4)²)

trigonometric substitution:
u=tan(z)

Hi Toutatis!

Try u = 2tan(z).

By Toutatis!

 

[Toutatis]

Hi Toutatis!

Try u = 2tan(z).

By Toutatis!

of course!

Thank you!