So I'm trying to integrate the indefinite integral of (2x/(x^2 + 6x + 13)) and I get to an answer that I know doesn't work but I'm only off but a very small amount and I'd like to know where I went wrong. I'll post my work and the incorrect answer below:
indefinite integral of ( (2x/(x^2 + 6x + 13)) dx)
indefinite integral of ( (2x/((x+3)^2 + 4)) dx)
Let u = x + 3
indefinite integral of ( (2(u-3)/(4 + u^2)) du)
(1/2) * indefinite integral of ( ((u-3)/(1 + (u/2)^2)) du)
(1/2) * ( indefinite integral of ( (u/(1 + (u/2)^2)) du) - indefinite integral of ( (3/(1 + (u/2)^2)) du) )
Let z = 1 + (u/2)^2
indefinite integral of (dz/z) - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)
ln( (x^2 + 6x + 13)/4 ) - (3/2) * arctan( (x+3)/2 ) + C
The correct answer should be:
ln(x^2 + 6x + 13) - 3*arctan( (x+3)/2 ) + C
indefinite integral of ( (2x/(x^2 + 6x + 13)) dx)
indefinite integral of ( (2x/((x+3)^2 + 4)) dx)
Let u = x + 3
indefinite integral of ( (2(u-3)/(4 + u^2)) du)
(1/2) * indefinite integral of ( ((u-3)/(1 + (u/2)^2)) du)
(1/2) * ( indefinite integral of ( (u/(1 + (u/2)^2)) du) - indefinite integral of ( (3/(1 + (u/2)^2)) du) )
Let z = 1 + (u/2)^2
indefinite integral of (dz/z) - (3/2) * indefinite integral of ( (1/(1 + (u/2)^2)) du)
ln( (x^2 + 6x + 13)/4 ) - (3/2) * arctan( (x+3)/2 ) + C
The correct answer should be:
ln(x^2 + 6x + 13) - 3*arctan( (x+3)/2 ) + C