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Let $\displaystyle I_n(x)=\int_{0}^{x}\frac{1}{(t^2+1)^n}\mbox{d}t$ for $n\in\mathbb{N}$. Find recursive formula for $I_n(x)$ that do not need integrals.

I don't know how to do such things and I have it in mind for a few days.

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    You'll need to do integration by parts with an appropriate splitting. What have you tried doing?2012-05-04

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Add and subtract $t^2$ from the numerator and you'll get $ I_n (x) = \int^{x}_0 \frac{1}{(t^2+1)^{n-1}} dt - \frac{1}{2}\int^x_0 \frac{(1+t^2)'\cdot t}{(1+t^2)^n} dt .$

I wrote the second term like that so you have a particularly easy application of integration by parts.

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    @wangdw Yes, the recurrence relation you obtained is incorrect.2012-05-04
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I might as well give my version of Ragib's fine answer. As I said, things become easier if you make the substitution $t=\tan\,u$. You end up with the integral

$\int_0^{\arctan\,x}\frac{\sec^2 u}{\sec^{2n}u}\mathrm du$

You can now try integration by parts on this integral; note that the trigonometric identity $\sec^2 u=1+\tan^2 u$ will be very helpful when evaluating compositions of the trigonometric function that pop up with the arctangent.

Don't hover below if you don't want to see the final answer:

$\int_0^x \frac{\mathrm dt}{\left(1+t^2\right)^{n+1}}=\frac{x}{2 n \left(1+x^2\right)^n}+\left(1-\frac1{2n}\right) \int_0^x \frac{\mathrm dt}{\left(1+t^2\right)^n}$

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    thank you for the final answer, I could check if I had good result.2012-05-04