Answer
$\dfrac{1}{6} \tan (3x^2+2x^3) +C$
Work Step by Step
Our aim is to solve the integral $ \int (x+x^2 ) \sec^2 (3x^2+2x^3) \ dx$
Let us consider that $a =3x^2+2x^3$ and $\dfrac{da}{dx}=6x+6x^2 \implies dx=\dfrac{1}{6(x+x^2 )} \ da$
Now, $ \int (x+x^2 ) \sec^2 (3x^2+2x^3) \ dx = \int (x+x^2 ) \sec^2 a (\dfrac{1}{6(x+x^2 )}) \ da=\dfrac{1}{6} \int \sec^2 a \ da$
or, $=\dfrac{1}{6} \tan a+C$
or, $=\dfrac{1}{6} \tan (3x^2+2x^3) +C$
