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