<s>[INST] Integrating $\int \frac{3}{(x^2 +5)^2}dx$ by parts Integrating $$\int \frac{3}{(x^2 +5)^2}dx$$
After removing the constant, it is basically integrating $\frac{1}{x^4+10x^2+25}$. I only have learnt up to integrating $\frac{1}{ax^2 + bx +c}$ with the highest power of $x$ is 2. And this cannot be broken up into partial fractions too. What happens when the power of $x$ is 4?
My thoughts are to do integration by parts (product rule)
$$\int u v' dx = uv - \int u' v dx$$
But I am unclear what do substitute $u$ and $v$ to
 [/INST] $$\int\frac3{2x}\cdot\frac{2x}{(x^2+5)^2}dx\\=\frac3{2x}\cdot\frac{-1}{(x^2+5)}-\int\frac{-3}{2x^2}\cdot\frac{-1}{x^2+5}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int\frac{x^2+5-x^2}{x^2(x^2+5)}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int(\frac1{x^2}-\frac1{x^2+5})dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10x}+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c\\=\frac{-3}{10x(x^2+5)}(5-x^2-5)+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c\\=\frac{3x}{10(x^2+5)}+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c$$
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