How do you integrate 1/(x^2 + 25)^2?

Question

I've been stuck on this question for a while. I've tried integration by parts but that doesn't help so I thought substitution would help but I don't know what to substitute it for.

Answer 1

Try the substitution x=5tanθ

Answer 2

From Michael's suggestion, dx = 5 sec²θ dθ.

 

1 / (x²+25)² = 1 / (25 tan²θ+25)² = cos⁴θ/25².

 

So, Int[1 / (x²+25)²] = Int[5 sec²θ cos⁴θ / 25²] = Int[cos²θ / 125] = Int[(1+cos(2θ)) / 250]

 

   = [θ+(1/2)sin(2θ)] /250

 

θ = tan^-1(x/5) gives (1/250) tan^-1(x/5)+(1/500) sin[2 tan^-1(x/5)]

 

Since, sin[2θ] =2 sinθ cosθ = 2 sin[tan^-1(x/5)] cos[tan^-1(x/5)] = 2 [x/√(x²+25)] [5/√(x²+25)] =10x / (x²+25), the integral yields

 

         (1/250) [tan^-1(x/5)+5x / (x²+25)]