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

How should I approach this? Is there a general approach when the degree of the denominator>>numerator in this case $x^4>>1$?

Pichi Wuana
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gbox
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    There is a general approach for all rational fractions. https://en.wikipedia.org/wiki/Partial_fraction_decomposition –  Feb 16 '16 at 20:25
  • @gbox see here someone asked the same today – 3SAT Feb 16 '16 at 20:41
  • $$\int\frac1{(x^2+1)^2},dx = \int\frac{1+x^2-x^2}{(x^2+1)^2},dx = \int\frac1{x^2+1},dx - \int x\frac{x}{(x^2+1)^2},dx$$$$ = \int\frac1{x^2+1},dx +\frac12\int x,d\frac1{x^2+1} = \int\frac1{x^2+1},dx +\frac12 \frac x{x^2+1} - \frac12\int\frac1{x^2+1},dx$$$$= \frac12\frac x{x^2+1} + \frac12\arctan x + const $$ – Yuri Negometyanov Feb 17 '16 at 09:02

2 Answers2

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Apply Integral Substitution: $\color{green}{x=\tan \left(u\right)\quad \:dx=\frac{1}{\cos ^2\left(u\right)}du}$ $$=\int \frac{\sec ^2\left(u\right)}{\left(\tan ^2\left(u\right)+1\right)^2}du=\int \frac{\sec ^2\left(u\right)}{\left(\sec ^2\left(u\right)\right)^2}du=\int \cos ^2\left(u\right)du=\int \frac{1+\cos \left(2u\right)}{2}du$$ Now it's easy...

Amarildo
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Here you can split the integrand into a sum of rational functions of the form $$\frac{1}{(x^2+1)^2}=\frac{a}{i+x}+\frac{b}{-i+x}+\frac{p}{(i+x)^2}+\frac{q}{(-i+x)^2},$$ determine the constants $a,\,b,\,p,\,q$ and then integrate.

TZakrevskiy
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