find the indefinite integral

Question

∫dx/(x²+a²)²

Answer 1

Let x = atanθ. Then dx = asec²θdθ

So the integral becomes ∫ [asec²θ] / [a²tan²θ + a²]² = ∫ [asec²θ] / [a⁴(tan²θ+1)²]dθ = (1/a³)∫cos²θdθ =

(1/a³)∫ [(1+cos(2θ))/2]dθ = (1/(2a³)) [θ+ (1/2)sin2θ] + C = (1/(2a³))[θ+ sinθcosθ]

Since tanθ = x/2, θ = Tan^-1(x/2), sinθ = x / √(x²+4) and cosθ = 2/√(x²+4)

So, we have (1/(2a³)) [Tan^-1(x/2) + 2x / (x²+4)] + C

Answer 2

This is a nuisance to integrate..but I will give you the steps.

Start with integrand 1/(x²+a²) and integrate it by parts with u=1/(x²+a²) and v=x so that dv=dx and

du=-2x/(x²+a²)²

Write out the integration by parts and you will find that you need to integrate 2x²/(x²+a²)² which is done by partial fractions with denominators (x²+a²) and (x²+a²)²

This will give you an equation including the integral you wanted in the first place^..

The result will be a rational fraction plus an arctan.