Answer
$\dfrac{1}{3} \ln |y^3-2|= x^3+c$
Work Step by Step
Rewrite the given equation and then integrate.
We have $ \int 3x^2 dx =\int \dfrac{y^2}{(y^3-2)}dy$
This can also be written as:
$\int 3x^2 dx =(\dfrac{1}{3}) \int \dfrac{3y^2}{(y^3-2)}dy$
Now, $x^3+c =\dfrac{1}{3} \int \dfrac{3y^2}{(y^3-2)}dy$
Suppose $a=y^3 \implies da =3y^2 dy$
$x^3+c =\dfrac{1}{3} \int \dfrac{3y^2}{(y^3-2)}dy \implies \dfrac{1}{3} \int \dfrac{da}{a-2}=x^3+c \implies \dfrac{1}{3} \ln |a-2|= x^3+c$
Hence, $\dfrac{1}{3} \ln |y^3-2|= x^3+c$
