Answer
$x \tan^{-1}(3x)- (\dfrac{1}{6}) \ln (9x^2+1)+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
$\int \tan^{-1}(3x dx)=x \tan^{-1}(3x)-\int (x) (\dfrac{3}{9x^2+1}) dx$
Thus, $x \tan^{-1}(3x)-\int (x) (\dfrac{3}{9x^2+1}) dx=x \tan^{-1}(3x)-\int (\dfrac{3x}{9x^2+1}) dx$
Plug $a=9x^2+1 \implies da=18xdx$
Then, we have $x \tan^{-1}(3x)-\int (\dfrac{3x}{9x^2+1}) dx=x \tan^{-1}(3x)-\int (\dfrac{1}{6a}) da=x \tan^{-1}(3x)-(\dfrac{1}{6}) \ln |a|+c'$
or, $x \tan^{-1}(3x)-\int (\dfrac{1}{6}) \ln |a|+c=x \tan^{-1}(3x)- (\dfrac{1}{6}) \ln (9x^2+1)+c$
