$\int \frac{dx}{\left(x^2+a^2\right)^3}$. I tried to use partial method were $u\:=\frac{1}{\left(x^2+a^2\right)^3}$ and dv = dx but got no result.
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4Hint: Substitute $x = a \tan(\theta)$. – Math Lover Sep 05 '17 at 19:52
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It's a trigonometric substitution and I'm surprised they haven't taught you that in baby calculus. That's a standard calculation method, it's nothing weird or exotic. Sigh-what are they teaching these kids now? – Mathemagician1234 Sep 05 '17 at 21:02
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Here's another way: https://math.stackexchange.com/a/689932/1242 – Hans Lundmark Sep 06 '17 at 06:54
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Use the substitution $x=a \tan{t}.$ Then $dx=\frac{a\,dt}{\cos^2{t}}, $ and $$\int \frac{dx}{\left(x^2+a^2\right)^3}=\int \frac{a\,dt}{\cos^2{t}\cdot a^6 \left(1+ \tan^2{t}\right)^3} = \frac{1}{a^5}\int \frac{\cos^6{t}\,dt}{\cos^2{t}}=\frac{1}{a^5}\int {\cos^4{t}\,dt}.$$
M. Strochyk
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Yup. And now the final integral is straightforward if not easy. – Mathemagician1234 Sep 05 '17 at 21:03