Question

Evaluate `int tan^-1x  dx`

Answer 1

Let I = `int tan^-1x  dx`

= `int tan^-1x.1 dx`

Integrating by parts

I = `tan^-1x int1dx - int(int1dx d/dx tan^-1x)dx`

= `tan^-1x(x) - intx*1/(1 + x^2)dx`

= `xtan^-1x - 1/2 int(2x)/(1 + x^2)dx`

= `xtan^-1x - 1/2log(1 + x^2) + c`