Answer
$x [\tan^{-1}(3x)]- (\dfrac{1}{6}) \ln (9x^2+1)+c$
Work Step by Step
The integration by parts formula suggests that $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$
Now, we have
$I=(x)( \tan^{-1}(3x))-\int x (\dfrac{3}{9x^2+1}) dx=x \tan^{-1}(3x)-\int (\dfrac{3x}{9x^2+1}) dx$
Consider $p=9x^2+1 $ or, $dp=18xdx$
$I=x \tan^{-1}(3x)-\int (\dfrac{1}{6p}) dp=x \tan^{-1}(3x)-(\dfrac{1}{6}) \ln |p|+c'$
$\implies I=x [\tan^{-1}(3x)] -\int (\dfrac{1}{6}) \ln |p|+c$
Thus, $I=x [\tan^{-1}(3x)]- (\dfrac{1}{6}) \ln (9x^2+1)+c$
