What is the integral of $1/((1+x^2)^n)$?
I have tried adding and substracting $x^2$ from the numerator
$$ \int \frac{1}{(1+x^2)^2}\:dx = \int \frac{1}{(1+x^2)^{n-1}}\,dx − \int \frac{x^2}{(1+x^2)^n}\,dx $$
but can't see how to proceed from here.
Asked
Active
Viewed 1,255 times
2
-
Also Reduction formula for$\int\frac{dx}{(1+x^2)^n}$ by $x=\tan u$ – Nov 23 '15 at 07:26
-
Sketch: Let $I_n$ be our integral. We obtain a reduction formula, by finding $I_{n-1}$ using integration by parts, $du=dx$ and $v=\frac{1}{(1+x^2)^{n-1}}$. Your trick will come up. – André Nicolas Nov 23 '15 at 07:29
-
Take x = tan y. The problem is reduced to finding the integral of cos y. – Saikat Feb 08 '16 at 23:05
-
dx = sec^2(y) dy = (1+ tan^2(y)) dy – Saikat Feb 08 '16 at 23:07