Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula?

Question

To calculate the antiderivate of $$\frac{1}{(x^2+1)^2}$$ , we can either use the iteration formula reducing the exercise to the integral of $\ \frac{1}{x^2+1}\ $ or we can use $$(\frac{x}{x^2+1})'=\frac{1-x^2}{(x^2+1)^2}$$ , but I do not see how we can get the antiderivate of $\ \frac{1-x^2}{(x^2+1)^2}\ $ either without any guess (If we know this antiderivate , we can express $\frac{1}{(x^2+1)^2}$ as a linear combination of $\ f(x):=\frac{1}{x^2+1}\ $ and $\ g(x):=\frac{1-x^2}{(x^2+1)^2}\ $ , namely $\ \frac{1}{(x^2+1)^2}=\frac{f(x)+g(x)}{2}\ $)

How can I calculate $\int \frac{dx}{(x^2+1)^2}$ only by using intgration by parts and the substitution rule as well as other basic integration rules ? I am looking for a solution not containing a guess or the iteration formula.

Then you can use $\dfrac{1}{x^2+1} = \dfrac{1}{2i}\biggl(\dfrac{1}{x-i} - \dfrac{1}{x+i}\biggr)$, square it, integrate, combine. Though Olivier Oloa's substitution is admittedly neater. — Daniel Fischer, Nov 22 '16 at 14:36

Olivier Oloa · Answer 1 · 2016-11-22T14:50:43.893

4

By the change of variable $$ x=\tan t, \quad dt= \frac{dx}{1+x^2}, $$ one gets $$ \int\frac{dx}{(1+x^2)^2}=\int\cos^2 t\:dt=\int \left(\frac12+\frac{\cos 2t}2\right)dt $$ which is easier.

edited Nov 22 '16 at 14:50

answered Nov 22 '16 at 14:34

Olivier Oloa

120,989

I think we have to ask the OP what he constitutes a guess. But +1 from me! – Chinny84 Nov 22 '16 at 14:35
@OlivierOloa It remains to transform $\sin(2\arctan(x))$. Do you mind to work that out as well ? (I think it is not too difficult, but it would be very nice if you added that) – Peter Nov 22 '16 at 14:53
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@Peter Yes, one may recall that $\sin(2u)=\frac{2u}{1+u^2}$ with $u=\arctan x$. – Olivier Oloa Nov 22 '16 at 14:56
OK, this is sufficient for me for an upvote :) – Peter Nov 22 '16 at 14:57
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@Peter Thank you very much for your feedback. – Olivier Oloa Nov 22 '16 at 14:58

score 3 · Accepted Answer · answered Nov 22 '16 at 14:39

We have

$$\int\frac{dx}{(1+x^2)^2}=\int \frac{1+x^2-x^2}{(1+x^2)^2}dx=$$

$$\arctan(x)+\frac{1}{2}\int x\frac{-2x}{(1+x^2)^2}dx=$$

$$\arctan(x)+\frac{1}{2}\left(\left[x \frac{1}{1+x^2}\right]-\int \frac{1}{1+x^2}\right)=$$

$$\frac{1}{2}\arctan(x)+\frac{x}{ 2(1+x^2) }+C$$

score 0 · Answer 3 · edited Apr 13 '17 at 12:21

For substitution method, see How do I solve this trigonometric substitution integral?.

For intgration by parts,

$$\int\dfrac{dx}{(1+x^2)^2}=\int\dfrac{2x}{(1+x^2)^2}\cdot\dfrac1{2x}\ dx$$

$$=\dfrac1{2x}\int\dfrac{2x}{(1+x^2)^2}dx-\int\left(\int\dfrac{2x}{(1+x^2)^2}dx\cdot \dfrac{d(1/2x)}{dx}\right)dx$$

$$=-\dfrac1{2x(1+x^2)}-\dfrac12\int\dfrac{dx}{(1+x^2)x^2}$$

Now $$\int\dfrac{dx}{(1+x^2)x^2}=\int\dfrac{(1+x^2)-x^2}{(1+x^2)x^2}dx=?$$

Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula?

3 Answers3