Answer
$(\dfrac{1}{2}) \sin^{-1} (\dfrac{x^2+1}{3}) +C $
Work Step by Step
We integrate the integral as follows:
$I=\int \dfrac{x dx}{\sqrt {9-(x^2+1)^2}}$
Using the Substitution Method, we get:
$x^2+1=u \implies du=2x$
Now, $I=(\dfrac{1}{2}) \times \int \dfrac{du}{\sqrt {9(1-u^2/9)}} \\=(\dfrac{1}{2}) \sin^{-1} (u/3) +C \\=(\dfrac{1}{2}) \sin^{-1} (\dfrac{x^2+1}{3}) +C $
