I worked differential linear equation and at end of equation I got this integral.Can someone give me a hint to do this: $$\int \frac{1}{\left(u^2+1\right)^2}du$$
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- $\begingroup$ See here (no trig substitutions!): math.stackexchange.com/a/689932/1242 $\endgroup$Hans Lundmark– Hans Lundmark2016-04-12 15:06:06 +00:00Commented Apr 12, 2016 at 15:06
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3 Answers
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1 Substitute $u=\tan x$ and $du=\sec^2 x dx$, so $(u^2+1)^2=(\tan^2x+1)^2=\sec^4 x$ and $x=\arctan u$
$$=\int \cos^2 x dx$$
Now write $\cos^2 x$ as $\frac 1 2 \cos 2x+\frac 1 2$
$$=\int\left(\frac 1 2 \cos 2x+\frac 1 2\right)dx=\frac{\sin 2x}{4}+\frac x 2+C$$
Substitute back $x=\arctan u$
$$=\color{red}{\frac{u^2\arctan u+u+\arctan u}{2u^2+2}+C}$$
- 1$\begingroup$ I want to do it without secant function,do you know another way to do that? $\endgroup$VidaXpo– VidaXpo2016-04-12 13:55:33 +00:00Commented Apr 12, 2016 at 13:55
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The trig substitution $u = \tan(x)$ is probably the best way to integrate this, but you can also integrate this by rewriting the integral as $$\int\frac{du}{(1+u^2)^2} = \int\Big(\frac{1}{1+u^2} + \frac{-u^2}{(1+u^2)^2}\Big)du$$ Edit: The second integral looked solvable without trig substitutions when I first typed it, but after looking at it again, a trig substitution looks unavoidable.
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Substitue $u = \tan x$
Then it reduces to
$\int \cos^2 x \:\text{dx}$
