integrate:xtan^-1/(1+x^2)^2....working plz

Answer 1

Not sure how to interpret

tan^-1/(1+x^2)^2

You don't seem to have a numerator for the fraction.

One way, which provides the easiest integration is

∫ x [tan(1+x^2)]^2 dx
Let
u = 1+x^2
du = 2x dx

and you have

1/2 ∫ tan^2(u) du
= 1/2 ∫ (sec^2(u) -1) du
= 1/2 (tan(u) - u)
= 1/2 (tan(1+x^2) - (1+x^2)) + C

If you meant inverse tan, try using arctan(...) to make it clear just what the argument is.