Question

Evaluate : `∫x  cot^-1x` dx

Answer 1

Let I = `∫x . cot^-1x` dx

I = `∫cot^-1x . x` dx

Using integration by parts , 

I = `cot^-1x ∫ x dx - ∫[d/dx(cot^-1 x)∫x  dx]dx`

∴ `I = cot^-1x . x^2/2 - ∫(-1)/(1 + x^2) . x^2/2 dx`

∴ I = `cot^-1x . x^2/2 + 1/2∫x^2/(1 + x^2) dx`

∴ I = `cot^-1x . x^2/2 + 1/2∫((1 + x^2) - 1)/(1 + x^2)`dx

∴ I = `cot^-1x . x^2/2 + 1/2∫(1 - 1/(1 + x^2))dx`

∴ I = `cot^-1x . x^2/2 + 1/2(x - tan^-1x) + "C"`

∴ I = `x^2/2 . cot^-1x + x/2 - 1/2 tan^-1x + "C"`