Answer
$\dfrac{1}{8} \ln (25+4x^2) +C$
Work Step by Step
Let us consider $u=25 +4x^2 \implies du =(8x) dx$
Now, the given integral can be written as:
$\dfrac{1}{8}\int\dfrac{1}{u} du=\dfrac{1}{8} \ln |u|+C $
or, $=\dfrac{1}{8} \ln (25+4x^2) +C$
