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
 A: Integrate by parts as follows
\begin{align}
\int \frac{3}{(x^2 +5)^2}dx &=\int \frac3{10x}d\left(\frac{x^2}{x^2+5}\right)\\
&= \frac{3x}{10(x^2+5)}+\frac3{10}\int\frac{1}{x^2+5}dx\\
 &= \frac{3x}{10(x^2+5)}+\frac3{10\sqrt5}\tan^{-1}\frac x{\sqrt5}+C
\end{align}
A: $$\frac{3}{10}\int\frac{10}{x^4+10x^2+25}dx$$
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, $$I_2=\frac3{10}\int\frac{dt}{t^2}\\=\frac{-3}{10t}+c_2\\=\frac{-3x}{10(x^2+5)}+c_2$$
A: Using partial fractions...
Replace $x=t\sqrt5$ and your integral becomes:
$$\frac{3\sqrt5}{25}\int \frac{dt}{(t^2+1)^2}$$
Start by writing:
$$\frac1{t^2+1}=\frac a{t+i}+\frac b{t-i}$$ and solving for $a,b.$ $a,b$ will be complex (they will be complex conjugates.)
Squaring, you get:
$$\frac1{(t^2+1)^2}=\frac{a^2}{(t+i)^2}+\frac{2ab}{t^2+1}+\frac{b^2}{(t-i)^2}$$
Integrating, you get: $$-\frac{a^2}{t+i}-\frac{b^2}{t-i}+2ab\arctan(t).$$
You can combine those first two terms to get something of the form $\frac{ct+d}{t^2+1},$ for $c,d$ real, and $2ab$ are real, too.
We can skip all that, then and take the derivative of $\frac{ct+d}{t^2+1}+f\arctan(t)$ to get: $$\frac{(c+f)(t^2+1)-2t(ct+d)}{(t^2+1)^2}$$
Setting the numerator to $1,$ we get $c+f-2c=0,$ $-2d=0,$ and $c+f=1.$ That means $c=1/2,d=0,f=1/2.$
So your integral should be $$\frac{3\sqrt5}{50}\left(\frac{t}{t^2+1}+\arctan(t)\right)$$ with $t=x/\sqrt5.$

This becomes harder if you have $x^4$ in place od $x^2.$ It's still doable with partial fractions, but the first trick of doing a simpler partial fraction then square isn't really possible, and it is also not possible the do the trick at the end where I avoided the complex numbers, so you'll need to know a thing or two about the logarithms of complex numbers.
But if the integral is of $\frac1{(t^2+1)^k}$ for some other $k,$ then you get an integral of the form $$\frac{p(x)}{(t^2+1)^{k-1}}+c\arctan(t)$$ for some polynomial $p(x)$ of degree less than $2k-2$ and constant $c.$
A: $$\int\frac3{2x}\cdot\frac{2x}{(x^2+5)^2}dx\\=\frac3{2x}\cdot\frac{-1}{(x^2+5)}-\int\frac{-3}{2x^2}\cdot\frac{-1}{x^2+5}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int\frac{x^2+5-x^2}{x^2(x^2+5)}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int(\frac1{x^2}-\frac1{x^2+5})dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10x}+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c\\=\frac{-3}{10x(x^2+5)}(5-x^2-5)+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c\\=\frac{3x}{10(x^2+5)}+\frac3{10\sqrt5}\tan^{-1}\frac{x}{\sqrt5}+c$$
A: Put $x=\sqrt5\tan\theta$, thus,
$$\int\frac{3\sqrt5\sec^2\theta d\theta}{(5\tan^2\theta+5)^2}\\=\frac3{5\sqrt5}\int\frac1{\sec^2\theta}d\theta\\=\frac3{10\sqrt5}\int2\cos^2\theta d\theta\\=\frac3{10\sqrt5}\int(1+\cos2\theta)d\theta\\=\frac3{10\sqrt5}(\theta+\frac{\sin2\theta}2)+c\\=\frac3{10\sqrt5}(\theta+\frac{\tan\theta}{1+\tan^2\theta})+c=\\\frac3{10\sqrt5}(\tan^{-1}\frac{x}{\sqrt5}+\frac{\frac{x}{\sqrt5}}{1+(\frac{x}{\sqrt5})^2})+c\\=\frac3{10\sqrt5}\left(\tan^{-1}\frac{x}{\sqrt5}+\frac{\sqrt5x}{5+x^2}\right)+c$$
