Answer
$$\int\frac{(x-2)^2\tan^{-1}(2x)-12x^3-3x}{(4x^2+1)(x-2)^2}dx=\frac{1}{4}(\tan^{-1}(2x))^2-3\ln|x-2|+\frac{6}{x-2}+C$$
Work Step by Step
$$A=\int\frac{(x-2)^2\tan^{-1}(2x)-12x^3-3x}{(4x^2+1)(x-2)^2}dx$$ $$A=\int\frac{(x-2)^2\tan^{-1}(2x)}{(4x^2+1)(x-2)^2}dx+\int\frac{-3x(4x^2+1)}{(4x^2+1)(x-2)^2}dx$$ $$A=\int\frac{\tan^{-1}(2x)}{4x^2+1}dx-3\int\frac{x}{(x-2)^2}dx$$ $$A=M-3N$$
1) Consider $M$:
$$M=\int\frac{\tan^{-1}(2x)}{4x^2+1}dx$$
Set $u=\tan^{-1}(2x)$, which means $$du=\frac{(2x)'}{(2x)^2+1^2}dx=\frac{2}{4x^2+1}dx$$ $$\frac{1}{4x^2+1}dx=\frac{1}{2}du$$
Therefore, $$M=\frac{1}{2}\int udu$$ $$M=\frac{1}{4}u^2+C$$ $$M=\frac{1}{4}(\tan^{-1}(2x))^2+C$$
2) Consider $N$:
$$N=\int\frac{x}{(x-2)^2}dx=\int\Big(\frac{A}{x-2}+\frac{B}{(x-2)^2}\Big)dx$$
We now clear the fractions.
$$A(x-2)+B=x$$ $$Ax-2A+B=x$$
So, $A=1$ while $-2A+B=0$, meaning that $B=2A=2\times1=2$
Therefore, $$N=\int\frac{dx}{x-2}+\int\frac{2}{(x-2)^2}dx$$ $$N=\ln|x-2|-\frac{2}{x-2}+C$$
So, in conclusion, $$A=M-3N=\frac{1}{4}(\tan^{-1}(2x))^2-3\ln|x-2|+\frac{6}{x-2}+C$$
