What is the integral of 1/((121x^2 + 4)^2) ?

Question

Im really not sure where to start with this. Thank you

Answer 1

I did the same substitution but after the substitution, I got 2/176∫1(tan²u + 1)² *sec²u du then use sec²u = 1 + tan²u to cancel out the secants to get 2/176∫1/sec²u du = 2/176∫cos²u du. At this point, put the cos²u + sin²u = 1 together with cos(2u) = cos²u - sin²u to get cos²u = 1/2 + 1/2cos(2u) and then sub that in to get: 2/176∫(1/2 + 1/2cos2u)du then you can integrate to get 1/176(sin(u)cos(u)) Then substitute back in u = arctan(11x/2) and sin(u) = 11x/sqrt(121x²+4) and cos(u) = 2/sqrt(121x²+4)

Ultimate answer: 1/176arctan(11x/2) + 1/8(x/(121x²+4) + C

Answer 2

∫ [1 / (121x² + 4)² ] dx = ∫ [1 / ((11x)² + 2²)²]dx

Let 11x = 2Tanu. Then dx = (2/11)Sec²udu

So, the given integral can be rewritten as (2/11)∫ [ Sec²u / (4Tan²u + 4)²]du

= (1/88)∫ [Sec²u / (Tan²u + 1)² ]du = (1/88)∫ [Sec²u / Sec⁴u]du

= (1/88)∫Cos²u du = (1/88)∫[(1+cos(2u))/2]du

= (1/178)[u + (1/2)sin(2u)] + C

= (1/178)[u + cosusinu] + C

Since tanu = 11x/2, cosu = 2/√(121x²+4) , sinu = 11x / √(121x²+4), and u = Arctan(11x/2)

So, (1/178)[u + cosusinu] + C = (1/178)[Arctan(11x/2) + 22x / (121x² + 4)] + C