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I need to derive a recursive formula for the integral $$ \int\frac{1}{(1+x^2)^k} dx\qquad(1)$$

I started from the assumption that $$x=tan(t) \qquad(2)$$ $$dx=sec^2(t) dt \qquad(3)$$ $$\frac{dx}{sec^2(t)}=dt$$

Subtitiution (2) into $1+x^2$:$$1+tan^2(t) = sec^2(t)\qquad(4)$$

Insert (3), (4) into (1)

$$I_k=\int\frac{1}{(1+x^2)^k} dx= \int\frac{1}{(sec^2(t))^k} \cdot sec^2(t) dt= \int (sec(t))^{-(2k-2)}dt = \int(cost)^{2k-2}dt$$

I know that I should apply integration by part to the element $$\int(cost)^{2k-2}dt$$but I don't know how to do it. I know the formula for this integration, but my tries bring crazy results. How to solve it?

Vid
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1 Answers1

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Let $I_k$ be your integral. Then, for $k>1$, $$ I_k=\int \frac{1+x^2-x^2}{(1+x^2)^k}\,dx= I_{k-1}-\int\frac{x^2}{(1+x^2)^k}\,dx= I_{k-1}+\frac{1}{2}\int x\frac{-2x}{(1+x^2)^k}\,dx $$ Now we can do by parts: $$ \int x\frac{-2x}{(1+x^2)^k}\,dx= x\frac{1}{k-1}\frac{1}{(1+x^2)^{k-1}}- \frac{1}{k-1}\int\frac{1}{(1+x^2)^{k-1}}\,dx $$

egreg
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