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How would one integrate the following?

$$\int \frac{x^{n-2}}{(1 + x)^n} {\rm d}x~$$ where $n$ is a positive integer.

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Hints: integration by parts and recursion over $n$. – Did Jun 18 '11 at 11:24
It's not meant to be a recursive/reduction formula though? – amy Jun 18 '11 at 11:30
I do not understand your comment. // Different hint: use some (clever) change of variable. – Did Jun 18 '11 at 11:33
By the way: what do you know, what have you tried, where are you stuck? And all this sort of things... – Did Jun 18 '11 at 11:34
I have tried the substitution $x = \tan^2 (\theta)$ and attempted integration by parts but I couldn't get any further on it. Which parts would I take if I was doing integration by parts? – amy Jun 18 '11 at 11:50
up vote 9 down vote accepted

$$\int {x^{n-2} \over (1+x)^n} {\rm d} x = \int (1+x)^{-2} \left({x \over 1+x}\right)^{n-2} {\rm d} x =\int y^{n-2} {\rm d} y$$

using the substitution $y = {x \over 1 + x}$.

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Thanks, your method was clever, but I don't think I'd ever have thought of that. – amy Jun 18 '11 at 12:00
@Jaydon: thanks. And by the way: me neither, up until now :) The main point is that you want to get rid of annoying terms like $(x / (1+x))^n$ and transform them into something simple that you know how to deal with. By using substitution we might have produced a complicated expression due to differentiation (indeed, that the result here is so simple is just a coincidence) but at least we isolate the hard power part into one simple $y^n$ term. – Marek Jun 18 '11 at 12:10

For me, the most natural way to begin is by letting $t=1+x$. Then the integral becomes $$ \int \frac{(t-1)^{n-2}}{t^n} dt = \int \left(1 - \frac{1}{t} \right)^{n-2} \frac{dt}{t^2}. $$ Then let $y=1/t$. Or maybe even better: $y=1-1/t$, which brings you back to Marek's answer.

If you would have had some other powers that didn't interact as nicely as $n-2$ and $n$, say $(t-1)^a/t^b$, you could have expanded using the binomial theorem and integrated term by term.

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I would like to do it with the following hint:


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My first instinct would have been to substitute $x = \tan^2(\theta)$. $$I = \int \frac{x^{n-2}}{(1 + x)^n} {\rm d}x$$ $x = \tan^2(\theta)$. \begin{align} I & = \int \dfrac{\tan^{2n-4}}{\sec^{2n}(\theta)} 2 \tan(\theta) \sec^2(\theta) d \theta\\ & = 2 \int \dfrac{\tan^{2n-3}}{\sec^{2n-2}(\theta)} d \theta\\ & = 2 \int \sin^{2n-3}(\theta)\cos(\theta) d \theta \end{align} Setting $\sin(\theta) = t$, we get $$I = 2 \int t^{2n-3} dt = 2 \dfrac{t^{2n-2}}{2n-2} + C = \dfrac{(\sin^2(\theta))^{n-1}}{n-1} + C = \dfrac1{n-1} \left(\dfrac{x}{1+x} \right)^{n-1} + C$$ This is just a round about way of doing what Marek did. But I would not have realized immediately to split it in such a nice way as Marek had done.

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