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I would like to know if there is an algebraic way to evaluate $$\int \frac{\text{d}x}{x^2(x^2+1)^2}$$ Without using partial fraction decomposition and without substitution.

My attempt is something like this $$\int \frac{\text{d}x}{x^2(x^2+1)^2}=\int \frac{(x^2+1)^2-x^4-2x^2}{x^2(x^2+1)^2}\text{d}x=\int \frac{1}{x^2}\text{d}x-\int \frac{x^2+2}{(x^2+1)^2}\text{d}x=$$ $$=-\frac{1}{x}+\int \frac{\text{d}x}{x^2+1}+\int \frac{1}{(x^2+1)^2}\text{d}x+c=-\frac{1}{x}+\arctan x+\int \frac{1}{(x^2+1)^2}\text{d}x+c$$

So I just need to integrate $\frac{1}{(x^2+1)^2}$, which is easy by letting $x=\tan \theta$; but actually I would like to end with algebra, so how can I integrate $\frac{1}{(x^2+1)^2}$ without substitution?

Thanks.

Bernkastel
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4 Answers4

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Use integration by parts

$$\int\dfrac{dx}{(1+x^2)^2}=\dfrac1x\int\dfrac{x\ dx}{(1+x^2)^2}-\int\left(\dfrac{d(1/x)}{dx}\int\dfrac{x\ dx}{(1+x^2)^2}\right)dx$$

Finally use $\dfrac1{x^2(x^2+1)}=\dfrac{x^2+1-x^2}{x^2(x^2+1)}=?$

lab bhattacharjee
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Firstly, evaluate :

$$\int\dfrac{dx}{(1+x^2)^2}$$

by substituting $x=\tan(\theta)$. Then you can verify that integral transforms to :

$$\int\cos^2{\theta} d\theta$$

Use the formula of double angles to evaluate this integral.

Finally,

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

First two terms are standard integrals and third term is the integral in the beginning of this answer.

Akash Yadav
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You can calculate the integral just by splitting it up several times as follows: $$\frac{1}{x^2(1+x^2)^2}= \frac 1{x^2(1+x^2)} - \frac 1{(1+x^2)^2}$$

So, you get $$\frac 1{x^2(1+x^2)} = \frac{1+x^2-x^2}{x^2(1+x^2)}=\frac 1{x^2}-\frac{1}{1+x^2}$$ and $$\frac 1{(1+x^2)^2} =\frac{1+x^2-x^2}{(1+x^2)^2} = \frac{1}{1+x^2} - \frac{x^2}{(1+x^2)^2}$$ (This step may answer your last question.)

Only one partial integration is left: $$\int\frac{x^2}{(1+x^2)^2}dx =\int x\frac{x}{(1+x^2)^2}dx=-\frac 12\frac{x}{1+x^2}+\frac 12\arctan x$$

Putting all this together:

$$\int \frac{\text{d}x}{x^2(x^2+1)^2} = -\frac 1x -\frac 32\arctan x -\frac 12 \frac x{1+x^2}(+C)$$

trancelocation
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Substitute $x=\tan(u)$ to get

\begin{align}\int\frac{\mathrm dx}{x^2(x^2+1)^2}&=\int\frac{\mathrm du}{\tan^2(u)\sec^2(u)}\\&=\int\frac{\cos^4(u)}{\sin^2(u)}~\mathrm du\\&=\int\frac{(1-\sin^2(u))^2}{\sin^2(u)}~\mathrm du\\&=\int\frac{1-2\sin^2(u)+\sin^4(u)}{\sin^2(u)}~\mathrm du\\&=\int\csc^2(u)-2+\sin^2(u)~\mathrm du\\&=-\cot(u)-\frac{3u}2-\frac{\sin(2u)}4+C\\&=-\frac1x-\frac32\arctan(x)-\frac x{2(x^2+1)}+C\end{align}

Simply Beautiful Art
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