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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? –  Guess who it is. May 4 '12 at 12:08

2 Answers 2

up vote 2 down vote accepted

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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@KannappanSampath Thanks for pointing that out!. –  Ragib Zaman May 4 '12 at 12:19
Anyway I am getting a stupid result that doesn't seem right: $$ \frac32I_n(x)=I_{n-1}(x)+I_{n+1}(x)-\frac12\frac x{(x^2+1)^{n-1}} $$ –  wangdw May 4 '12 at 12:23
Figuring out how this splitting is done might be slightly easier if one makes the preliminary substitution $t=\tan\,u$ –  Guess who it is. May 4 '12 at 12:36
@wangdw Yes, the recurrence relation you obtained is incorrect. –  Sasha May 4 '12 at 13:08

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. –  ray May 4 '12 at 15:18

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