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Integrate $\int \frac{1+t^2}{(1-t^2)^2} dt \\$ I just want some hint, not the full answer. I've tried substituting $v= 1\pm t^2$ and I've tried partial fraction but I end up with weird results. Any help?

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First write

$$\int \frac{1 + t^2}{(1 - t^2)^2} = \int \frac{(1 + t)^2 - 2t}{(1 - t^2)^2}\, dt = \int \frac{(1 + t)^2}{(1 - t^2)^2} + \int \frac{-2t\, dt}{(1 - t^2)^2}.$$

Since $1 - t^2 = (1 - t)(1 + t)$,

$$\int \frac{(1 + t)^2}{(1 - t^2)^2}\, dt = \int \frac{(1 + t)^2}{(1 - t)^2(1 + t)^2}\, dt = \int \frac{1}{(1 - t)^2}\, dt,$$

which can be evaluated by the $u$-substitution $u = 1 - t$. To evaluate $\int -2t/(1 - t^2)^2\, dt$, use the $v$-substitution $v = 1 - t^2$ and note that $dv = -2t\, dt$.

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Partial fraction decomposition works: $$\int \frac{1+t^2}{(1-t^2)^2} dt = \int \frac{1+t^2}{(t-1)^2(t+1)^2}\,dt= \int \left(\frac{A}{t+1} + \frac B{(t+1)^2} + \frac C{t-1} + \frac D{(t-1)^2}\right)\,dt$$

$$A(t+1)(t-1)^2 + B(t-1)^2 + C(t-1)(t+1)^2 + D(t+1)^2 = 1+t^2$$

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  • $\begingroup$ But $(1-t^2)^2=(1-t)(1+t)(1-t)(1+t)$ so shouldn't it be $\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1-t}+\frac{D}{1+t}$? If not, why? $\endgroup$ – Rousseau Feb 13 '15 at 21:46
  • $\begingroup$ @Rousseau That's not how partial-fractions are set up when you have a polynomial in the denominator which has for e.g. two 'repeated' roots. $\endgroup$ – Workaholic Feb 13 '15 at 21:49
  • $\begingroup$ Try reconstructing the original fraction using your proposal, finding common denominator. It won't work. In general, $$\int \frac 1{(x-1)^n}\,dx = \int \frac {A_1}{(x-1)} + \frac{A_2}{(x-1)^2} + \cdots + \frac{A_n}{(x-1)^n}\,dx$$ $\endgroup$ – Namaste Feb 13 '15 at 21:50
  • $\begingroup$ @amWhy Hm yeah I got the result 2=0 (hence 'weird results'). Didn't think of what Workaholic said about multiplicity. Just curious, how do you prove that if a polynomial has multiplicity n it must be decomposed into n factors from degree 1 to n? $\endgroup$ – Rousseau Feb 13 '15 at 21:55
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Hint:

$$ \begin{align*} \int \frac{1+t^2}{(1-t^2)^2}\,{\rm d}t &= \int \frac{1-t^2+2t^2}{(1-t^2)^2}\,{\rm d}t\\ &= \int\left[\frac{1-t^2}{(1-t^2)^2}+\frac{2t^2}{(1-t^2)^2}\right]{\rm d}t\\ &= \int\frac{1}{1-t^2}\,{\rm d}t+2\int\frac{t^2}{(1-t^2)^2}\,{\rm d}t. \end{align*} $$

Is it simpler now?

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  • $\begingroup$ I don't think it's easier to integrate $\frac{t^2}{1-t^2}^2$, but maybe I miss some neat method. Were you thinking partial fraction decomposition? $\endgroup$ – Rousseau Feb 13 '15 at 21:49
  • $\begingroup$ @Rousseau You can simplify it even more by applying the very same trick we used to get : $$\frac{t^2}{(1-t^2)^2} = \frac{t^2-1+1}{(1-t^2)^2} = \frac{t^2-1}{(1-t^2)^2}+\frac{1}{(1-t^2)^2}=\frac{1}{(1-t^2)^2}-\frac{1}{1-t^2}.$$ $\endgroup$ – Workaholic Feb 13 '15 at 21:56
  • $\begingroup$ Yes but I don't think it's much easier to integrate $\frac{1}{(1-t^2)^2}$ than the originial integral. The problem was the partial fraction decomposition and I guess there's no apparent simple way to get around that. $\endgroup$ – Rousseau Feb 13 '15 at 21:59
  • $\begingroup$ @Rousseau Yes, because you would have to decompose those fractions anyway. Here's the set-up: $$\dfrac1{(1-t^2)^2}=\dfrac{1}{(1-t)^2(1+t)^2}=\dfrac{A}{1-t}+\dfrac{B}{(1-t)^2}+\dfrac{C}{1+t}+\dfrac{D}{(1+t)^2}.$$ $\endgroup$ – Workaholic Feb 13 '15 at 22:03
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Maybe use

$\dfrac{1+t^2}{(1-t^2)^2} = \dfrac{1}{2}\dfrac { (1-t)^2 + (1+t)^2}{((1-t)(1+t))^2}$

so you end up integrating $\dfrac{1}{2}\cdot \left( \dfrac{1}{(1+t)^2} + \dfrac{1}{(1-t)^2} \right)$

where each is easy with substitution $1+t = u, 1-t = v$

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Try the substitution $t = \sin(x)$. This is a challenging integral no matter what you do. If you use this substitution you will have: $$\begin{align}\int\frac{1+\sin^2(x)}{(1-\sin^2(x))^2}\cos(x)dx = \int\frac{\cos(x)+\cos(x)\sin^2(x)}{\cos^4(x)}dx \\ = \int\frac{1}{\cos^3(x)}+\frac{\sin^2(x)}{\cos^3(x)}dx \\ = \int\sec^3(x)dx+\int\tan^2(x)\sec(x)dx \\ = \int\sec^3(x)dx+\int(\sec^2(x)-1)\sec(x)dx \\ = \int\sec^3(x)dx+\int\sec^3(x)dx-\int\sec(x)dx \\ = 2\int \sec^3(x) -\int \sec(x)dx \end{align}$$ It is well known (or can be found in an integral table) that $\int\sec(x)dx = \ln(\sec(x)+\tan(x))$. Hence $\int\sec^3(x)dx$ remains to be integrated for a final answer. I recommend integration by parts to calculate this integral, or looking it up in a table.

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  • $\begingroup$ $\frac{1+sin^2x}{cos^4x} = \frac{cos^2x}{cos^4x}+\frac{2sin^2x}{cos^4x}$. Is this the right track or have I derailed? $\endgroup$ – Rousseau Feb 13 '15 at 22:12
  • $\begingroup$ @Rousseau That is not incorrect! Just not the way I went about it. You should still get a result if you proceed. It may or may not be easier than what I did. I broke down my process a bit more, since you already chose an answer on your question anyway. $\endgroup$ – graydad Feb 13 '15 at 22:14
  • $\begingroup$ To be honest, I think your integral looks more frightening than the one I started with. I guess $sec^3(x) dx$ can be evaluated with the substitution $tan(x/2)=v$ and integration by parts. But I'm not sure it's easier than the original integral...? $\endgroup$ – Rousseau Feb 13 '15 at 22:18
  • $\begingroup$ @Rousseau I had made a mistake in my equation but I caught it. And actually $\int \sec^3(x)dx$ is surprisingly easy if you already know $\int \sec(x)dx$. It follows quickly from letting $dv = \sec^2(x)$, $u = \sec(x)$ and after integrating by parts the first time using $\tan^2(x) = \sec^2(x)-1$. Integrating by parts a second time is not required at this point. But I fully understand trig integrals looking frightening. Just thought I'd offer another approach to your question :) $\endgroup$ – graydad Feb 13 '15 at 22:20
  • $\begingroup$ Can you explain $dv=sec^2(x), u = sec(x)$? Are you introducing two new variables? $\endgroup$ – Rousseau Feb 13 '15 at 22:26

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