3

Can someone help me to compute: $$\int \frac{x}{(x^2-4x+8)^2}\mathrm dx$$ And, in general, the type:

$$\int \frac{N(x)}{(x^2+px+q)^n}\mathrm dx$$ with the order of polynomial $N(x)<n$ and $n$ natural greater than 1?

Ali Caglayan
  • 5,726
bleish
  • 109
  • possible duplicate of Integration by partial fractions; how and why does it work? – Peter Woolfitt Aug 10 '14 at 14:06
  • 1
    @Peter This question is not a duplicate, and does not necessarily require partial fraction decomposition. (Nor did the OP ask to use it.) Even if used, the methods below would be needed. Why use it when the methods below (needed in any case) suffice? – amWhy Aug 10 '14 at 14:28
  • @amWhy The OP asked for a more general case. The answers below don't answer that. I think it is a duplicate. – Ayman Hourieh Aug 10 '14 at 14:30
  • @AymanHourieh I agree with amWhy: This question is asking for a formula for the integral in terms of $N(x)$, $p$, $q$, and $n$. The other is asking for information about partial fractions. Not a duplicate. – apnorton Aug 10 '14 at 14:37
  • 1
    @amWhy as Ayman says, I marked this as a duplicate because of the second (quite general) part of the question. Of course the first part can be done with a variety of methods, but my understanding was that generality for this question comes with partial fraction decomposition (in particular covered under case 5 of the answer I linked to). Still, in retrospect it would have been better just to post a link to the other question and not to have proposed the duplicate, so I apologize. – Peter Woolfitt Aug 10 '14 at 14:43
  • @anorton Do you have an answer to the OP's second question that isn't a subset of the generalized answered posted by Peter? We have generalized answers for a reason. – Ayman Hourieh Aug 10 '14 at 14:44
  • @AymanHourieh I do not, but that doesn't mean that no one does. I don't know about complex methods, but I wonder if that could lead to further simplification. – apnorton Aug 10 '14 at 16:44
  • @anorton If there is a method not mentioned in the general answer, it should be added there. – Ayman Hourieh Aug 10 '14 at 17:02
  • @AymanHourieh some methods may work for this particular case that do not work for the prior. Insisting that all methods that work for this question will work for the generalized form is crazy. – apnorton Aug 10 '14 at 17:05
  • @anorton Let's keep it civil, shall we? All what I'm saying is that the answers presented here are merely special cases of a general solution we already have on this site. The reason we have a generalized answer is precisely to avoid a proliferation of special cases that don't add value. Given how straightforward it is to apply the general solution to this case, I doubt that there is value in waiting for new methods. – Ayman Hourieh Aug 10 '14 at 17:11

2 Answers2

5

$$\int \frac{x}{(x^2-4x+8)^2} dx = \int\frac{x - 2 + 2}{(x^2 - 4x + 8)^2}\,dx $$ $$= \frac 12\int \frac{2x - 4}{(x^2 - 4x + 8)^2}\,dx + \int \frac 2{((x-2)^2 + 2^2)^2}\,dx$$

For the first integral, use $u = x^2 - 4x + 8 \implies du = (2x-4)\,dx$.

For the second integral, put $2\tan \theta = (x-2)\implies 2\sec^2 \theta\,d\theta = dx$.

amWhy
  • 209,954
1

HINT:

As $\displaystyle x^2-4x+8=(x-2)^2+2^2,$

use Trigonometric substitution as $x-2=2\tan\theta$

lab bhattacharjee
  • 274,582
  • Nice hint lab. :) – Mikasa Aug 11 '14 at 02:21
  • During the computation, we have to integrate $\cos^2x$, i.e. we have to find the integral of $\sin x \cos x=\tan x \frac{1}{1+\tan^2x} $. Only now, we can subsitute the upper expression. – bleish Aug 11 '14 at 16:31
  • @bleish, $$\int\cos^2x\ dx=\int\frac{1+\cos2x}2dx=\frac x2+\frac{\sin2x}4+C$$ and $$\int\sin x\cos x\ dx=\int\frac{\sin2x}2=-\frac{\cos2x}4+K$$ – lab bhattacharjee Aug 11 '14 at 16:33
  • @labbhattacharjee: Ok for the first expression. but, the second? why another integration? – bleish Aug 11 '14 at 18:02
  • @bleish, $$y=\arctan\frac{x-2}2\implies \tan y=\frac{x-2}2,$$ We need to calculate $$\sin2y,\cos2y$$ See http://en.wikipedia.org/wiki/Tangent_half-angle_substitution – lab bhattacharjee Aug 11 '14 at 18:09
  • @labbhattacharjee: ok, ok. we only need to calculate $\sin 2y$ . $$\sin 2y=2 \tan y \frac{1}{1+ \tan^2 y}= \frac{x-2}{2} \frac{2}{1+ \frac{(x-2)^2}{4}}$$. There are no integral $\int \sin y \cos y dy$ – bleish Aug 11 '14 at 18:15
  • @bleish, $$\sin2y=\frac{2\tan y}{1+\tan^2y}$$ – lab bhattacharjee Aug 11 '14 at 18:18