Hard Integral without Partial Fraction Decomposition So I was trying to use partial fraction decomposition on this problem, and I realized that it didn't work, as it is already in partial fraction decomposition form.
$\int{\frac{3x+4}{(x^2+5)^2}dx}=$
Well, $\frac{3x+4}{(x^2+5)^2}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}$
$(x^2+5)^2[\frac{3x+4}{(x^2+5)^2}]=(x^2+5)^2[\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}]$
$3x+4=(Ax+B)(x^2+5)+(Cx+D)$
$3x+4=Ax^3+5Ax+Bx^2+5B+Cx+D$
$3x+4=(A)x^3+(B)x^2+(5A+C)x+(5B+D)$
So, $A=0$, $B=0$, $5A+C=3$, and $5B+D=4$
So, $A=0$, $B=0$, $C=3$, and $D=4$
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{(0)x+(0)}{x^2+5}+\frac{(3)x+(4)}{(x^2+5)^2}}dx$
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{3x+4}{(x^2+5)^2}}dx.$
As this clearly doesn't work, I am wondering what the integral would be. WolframAlpha tells me that the integral is as follows:
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\frac{1}{50}(\frac{5(4x-15)}{x^2+5}+4\sqrt{5}\tan^{-1}(\frac{x}{\sqrt{5}}))+c$
I am not sure how to get here. Any help would be greatly appreciated.

hint
Write it as the sum $ I+J$ with
$$I=\frac 32\int \frac{2xdx}{(x^2+5)^2}$$
easy
and
$$J=\frac 45\int \frac{x^2+5-x^2}{(x^2+5)^2}dx$$
$$=K+L$$
where $ K =\frac 45\int \frac{1}{x^2+5}dx$ is easy and
$$L=-\frac 25\int x \frac{2x}{(x^2+5)^2}dx$$
which we integrate by parts.
