Question

Evaluate the integral (tangent inverese of x/(x)^2)dx

Answer 1

Well if the question is integral of \[\tan^{-1} (x/x^{2})\]
Then that equals to integral of \[\tan^{-1} (1/x)\]
which by using the formula for integral of \[\tan^{-1} (x)\] =\[1/\sqrt{1+x^2}\]
here it will be \[1/\sqrt{1+(1/x^2)} = x/\sqrt{x^2+1}\]

Answer 2

no is tan^-1(x) over x^2

Answer 3

you dont know it ?

Answer 4

integral of [tan^-1(x)]/x^2 dx
=integral of (x^-2)[tan^-1(x)] dx
= -(x^-1)[tan^-1(x)] +ln[(x)/(1+x^2)]+C

Answer 5

simagholami do you have answer choices?