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What is $$\int \frac{4t}{1-t^4}dt$$ is there some kind of substitution which might help .Note that here $t=\tan(\theta)$

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Why not decompose the fraction instead? You would then get that, $$\dfrac{4t}{1-t^4}=\dfrac{2t}{t^2+1}-\dfrac{1}{t-1}-\dfrac{1}{t+1}.$$ – Workaholic Mar 9 at 16:02
Can someone tell me what $t= \tan \theta$ means? – MXYMXY Mar 9 at 16:07
Uh oh, I get the feeling that the integral he wants to do is $\int \frac{4\tan\theta}{1-\tan^4\theta}dt$. So all our answers are to the wrong question! – Mark Fischler Mar 9 at 16:24
@MarkFischler Oh.. I think the OP would need to clarify that. – MXYMXY Mar 9 at 16:25
@MarkFischler: $d \theta$ or $d \tan \theta$? Those are two very different integrals, and if you're right, it's certainly not $dt$. – Kevin Mar 10 at 5:21
up vote 8 down vote accepted

Substitution is not necessary. However, $u=t^2$ will be helpful.

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Thanks got it!!! Short and simple – Archis Welankar Mar 9 at 17:06
You are welcome. – André Nicolas Mar 9 at 17:12

Well aside the $4$ constant, you have

$$\int\frac{t}{1 - t^4}\ \text{d}t$$


$y = t^2$ so $\text{d}y = 2t\ \text{d}t$


$$2\int \frac{1}{1 - y^2}\ \text{d}y = 2\ \text{arctanh}(y) ~~~ \to ~~~ 2\ \text{arctanh}(t^2)$$

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Don't forget that you can also write the hyperbolic arc tangent in terms of logarithms! – Beta Mar 9 at 16:07

Observe that $$1-t^4=1-(t^2)^2$$

Let $$u=t^2$$ with $$du = 2t\,dt$$

The integral becomes

$$I=\int \frac{2}{1-u^2}du$$

Which is equal to

$$I = \ln{\frac{|u+1|}{|u-1|}}+C=\ln{\frac{|t^2+1|}{|t^2-1|}}+C$$

The above answer has a formula, but can be most conveniently reached using partial fractions method.

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Doesn't this work?

$$\int \frac{4t}{1-t^4}dt=\int \frac{2t}{1+t^2}-\frac{-2t}{1-t^2} dt=\ln |\frac{1+t^2}{1-t^2}|+C$$

Also $t=\tan \theta$ substitution works quite well. $$\int \frac{4\tan \theta}{1-\tan^4 \theta}\sec^2 \theta d \theta=\int \frac {4 \tan \theta }{1-\tan^2 \theta} d\theta=2\int \frac{1+\tan \theta}{1-\tan \theta}-\frac{1-\tan \theta}{1+\tan \theta} d\theta$$

$$2\int \frac{1+\tan \theta}{1-\tan \theta}-\frac{1-\tan \theta}{1+\tan \theta} d\theta=2\int\frac{cos \theta+\sin \theta}{\cos \theta-\sin \theta}+\frac{\cos \theta-\sin \theta}{cos \theta+\sin \theta}d\theta$$

Note that $(\sin \theta+\cos \theta)'=\cos \theta-\sin \theta$, $(\cos \theta-\sin \theta)'=-\sin \theta-\cos \theta$.

I think you can continue from here.

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HINT: we get $$\frac{4t}{1-t^4}=- \left( t-1 \right) ^{-1}- \left( t+1 \right) ^{-1}+2\,{\frac {t}{{t} ^{2}+1}} $$

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Hint: $t^2=u$ or $t^2=\sin v$

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$$I=\int \frac{4t}{1-t^4}dt$$ Let $t=\sqrt{\sin\theta}$. Then, $dt=\frac{1}{2\sqrt{\sin\theta}}\cos\theta d\theta$. Then, $$I=\int \frac{4\sqrt{\sin\theta}\cos\theta d\theta}{2\sqrt{\sin\theta}\cos^2\theta}=2\int\sec\theta d\theta=2\ln|\sec\theta+\tan\theta|+c=2\ln\left|\frac1{\cos\theta}+\frac{\sin\theta}{\cos\theta}\right|+c=2\ln\left|\frac{t^2+1}{\sqrt{1-t^4}}\right|+c=\ln\left|\frac{(t^2+1)^2}{(1+t^2)(1-t^2)}\right|+c=\ln\left|\frac{t^2+1}{1-t^2}\right|+c$$

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Couldn't you simplify yours further by putting the $2$ in? – MXYMXY Mar 9 at 16:24

The useful technique to find a good substitution is to let $t = f(u;a)$ where the function $f$ is of some simple form but contains a parameter $a$. In this case, the natural thing to try is to let $$ t = u^a. $$ Then $$ \int \frac{4t}{1-t^4}dt = \int \frac{4u^a}{1-u^{4a}} au^{a-1} du = \int\frac{4au^{2a-1}}{1-u^{4a}}du $$ Now you choose $a$ to make the integral simpler. Here, the two choices that come to mind are $a=\frac{1}{4}$ to simplify the denominator, or $a=\frac{1}{2}$ to simplify the numerator. The former leads to $\int \frac{\sqrt{u}}{-1u}$ which is not trivial. So go with $a=\frac{1}{2}$ to get $$ \int \frac{2}{1-u^2} du = \log(u+1) - \log(u-1) = \log(t^2+1) - \log(t^2-1)$$


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