Answer
$$\int\frac{(x+1)^2\tan^{-1}(3x)+9x^3+x}{(9x^2+1)(x+1)^2}dx=\frac{1}{6}(\tan^{-1}(3x))^2+\ln|x+1|+\frac{1}{x+1}+C$$
Work Step by Step
$$A=\int\frac{(x+1)^2\tan^{-1}(3x)+9x^3+x}{(9x^2+1)(x+1)^2}dx$$ $$A=\int\frac{(x+1)^2\tan^{-1}(3x)}{(9x^2+1)(x+1)^2}dx+\int\frac{x(9x^2+1)}{(9x^2+1)(x+1)^2}dx$$ $$A=\int\frac{\tan^{-1}(3x)}{9x^2+1}dx+\int\frac{x}{(x+1)^2}dx$$ $$A=M+N$$
1) Consider $M$:
$$M=\int\frac{\tan^{-1}(3x)}{9x^2+1}dx$$
Set $u=\tan^{-1}(3x)$, which means $$du=\frac{(3x)'}{(3x)^2+1^2}dx=\frac{3}{9x^2+1}dx$$ $$\frac{1}{9x^2+1}dx=\frac{1}{3}du$$
Therefore, $$M=\frac{1}{3}\int udu$$ $$M=\frac{1}{6}u^2+C$$ $$M=\frac{1}{6}(\tan^{-1}(3x))^2+C$$
2) Consider $N$:
$$N=\int\frac{x}{(x+1)^2}dx=\int\Big(\frac{A}{x+1}+\frac{B}{(x+1)^2}\Big)dx$$
We now clear the fractions.
$$A(x+1)+B=x$$ $$Ax+A+B=x$$
So, $A=1$ while $A+B=0$, meaning that $B=-A=-1$
Therefore, $$N=\int\frac{dx}{x+1}-\int\frac{dx}{(x+1)^2}$$ $$N=\ln|x+1|+\frac{1}{x+1}+C$$
So, in conclusion, $$A=M+N=\frac{1}{6}(\tan^{-1}(3x))^2+\ln|x+1|+\frac{1}{x+1}+C$$
