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