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

$$\frac{3}{10}\int\frac{10}{x^4+10x^2+25}dx$$
Dividing numerator and denominator by $x^2$
$$\frac{3}{10}\int\frac{\frac{10}{x^2}}{x^2+10+(\frac5x)^2}dx\\=\frac{3}{10}\int\frac{\frac5{x^2}+1+\frac5{x^2}-1}{x^2+10+(\frac5x)^2}dx\\=\frac{3}{10}\int\frac{1+\frac5{x^2}}{x^2+10+(\frac5x)^2}dx-\frac3{10}\int\frac{1-\frac5{x^2}}{x^2+10+(\frac5x)^2}dx\\=I_1-I_2$$
For $I_1$, put $x-\frac5x=t$. Its a) derivative is $(1+\frac5{x^2})dx=dt$ and b) square is $x^2-10+(\frac5x)^2=t^2$
Thus, $$I_1=\frac3{10}\int\frac{dt}{t^2+20}\\=\frac{3}{20\sqrt5}\tan^{-1}\frac{t}{2\sqrt5}+c_1\\=\frac3{20\sqrt5}\tan^{-1}\frac{x^2-5}{2\sqrt5x}+c_1$$
For $I_2$, put $x+\frac5x=t$. Its a) derivative is $(1-\frac5{x^2})dx=dt$ and b) square is $x^2+10+(\frac5x)^2=t^2$
Thus, $$I_2=\frac3{10}\int\frac{dt}{t^2}\\=\frac{-3}{10t}+c_2\\=\frac{-3x}{10(x^2+5)}+c_2$$
