Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would one integrate the following?

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

share|improve this question
    
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? –  j_z Jun 18 '11 at 11:30
1  
I do not understand your comment. // Different hint: use some (clever) change of variable. –  Did Jun 18 '11 at 11:33
2  
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? –  j_z Jun 18 '11 at 11:50

4 Answers 4

up vote 8 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 $y = {x \over 1 + x}$.

share|improve this answer
    
Thanks, your method was clever, but I don't think I'd ever have thought of that. –  j_z 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

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.

share|improve this answer

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.

share|improve this answer

I would like to do it with the following hint:

$$\int\frac{x^{n-2}}{(1+x)^n}dx=\int\frac{x^{n-2}}{(1+x)^2(1+x)^{n-2}}dx=\int\frac{1}{(1+x)^2}(1-\frac{1}{1+x})^{n-2}dx$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.