Answer
$-\dfrac{\sqrt {x^2+1} }{x}+C$
Work Step by Step
Let us consider $x = \tan \theta $
and $dx= \sec^2 \theta d\theta $ and $\sec \theta =\sqrt {x^2+1}$
Now, the given integral becomes:
$\int \dfrac{ \sec^2 \theta }{\tan^2 \theta( \sec \theta)} d \theta = \int \dfrac{\cos \theta}{\sin^2 \theta} d \theta$
Let us substitute $u=\sin \theta \implies du =\cos \theta d \theta$
we have $=\dfrac{1}{u^2} du$
Now, integrate:
$\dfrac{-1}{u}+C=\dfrac{-1}{\sin \theta}+C=-\csc \theta +C$
and plug in $\sec \theta =\sqrt {x^2+1} \implies \csc \theta =\dfrac{\sqrt {x^2+1} }{x}$:
Thus, $=-\dfrac{\sqrt {x^2+1} }{x}+C$
