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?


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 $dv = -2t\, dt$.


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

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

  • $\begingroup$ 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


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

Is it simpler now?

  • $\begingroup$ 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

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 $1+t = u, 1-t = v$


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.

  • $\begingroup$ $\frac{1+sin^2x}{cos^4x} = \frac{cos^2x}{cos^4x}+\frac{2sin^2x}{cos^4x}$. 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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