Question

Evaluate the following : `int x^2tan^-1x.dx`

Answer 1

Let I = `int x^2 tan^-1 x.dx`

= `int(tan^-1x).x^2dx`

= `(tan^-1x) int x^2.dx - int[{d/dx(tan^-1x) int x^2.dx}].dx`

= `(tan^-1 x)(x^3/3) - int (1/(1 + x^2))(x^3/3).dx`

= `x3/(3) tan^-1x - (1)/(3) (x(x^2 + 1) - x)/(x^2 + 1).dx`

= `x^3/(3) tan^-1x - (1)/(3)[int{x - x/(x^2 + 1)}.dx]`

= `x^3/(3) tan^-1x - (1)/(3)[int x.dx - (1)/(2) int(2x)/(x^2 + 1).dx]`

= `x^3/(3)tan^-1x - (1)/(3) [x^2/(2) - (1)/(2)log|x^2 + 1|] + c`

...`[because d/dx(x^2 + 1) = 2x and int (f'(x))/f(x) dx = log|f(x)| + c]`

= `x^3/(3)tan^-1x - x^2/(6) + (1)/(6) log|x^2 + 1| + c`.