Integration of Inverse Trig Functions?

Let INT = integral symbol

INT (2x)/(x^2+6x+13)

x² + 6x + 13 = (x+3)² + 4

Use trig substitution:
x + 3 = 2 tanu
dx = 2 sec²u du

∫ 2x/(x²+6x+13) dx = ∫ 2x/((x+3)²+4) dx
.......................... = ∫ 2(2 tanu − 3) / (4 tan²u+4) * 2 sec²u du
.......................... = ∫ 2(2 tanu − 3) / (4 sec²u) * 2 sec²u du
.......................... = ∫ (2 tanu − 3) du
.......................... = −2 ln |cosu| − 3u + c

Substitute back: tanu = (x+3)/2 ----> cosu = 2/√((x+3)²+2²) = 2/√(x²+6x+13)

.......................... = −2 ln| 2/√(x²+6x+13)| − 3tan⁻¹((x+3)/2) + c
.......................... = ln |(x²+6x+13)/4| − 3tan⁻¹((x+3)/2) + c
.......................... = ln (x²+6x+13) − 3tan⁻¹((x+3)/2) + (c − ln4)
.......................... = ln (x²+6x+13) − 3tan⁻¹((x+3)/2) + C

step by step solution:
http://symbolab.com/solver/integral-calc...

hope this helps