Understanding substitution to compute primitive To compute the primitive
$$\int x^2 \sqrt[3]{x^3+3}\ dx$$
I am trying this:
$t = x^3+3$
$dt = 3x^2dx$
but
$\int x^2 \sqrt[3]{x^3+3}\ dx=\ ?$
How to continue from here?
 A: I just want to make it clear since the OP looks like didn't understand yet that the answer provided by Chris K indeed using substitution method. Your attempt is correct by letting $t=x^3+3$ and $dt=3x^2\ dx\;\Rightarrow\;dx=\dfrac{dt}{3x^2}$, then the integral turns out to be
$$
\begin{align}
\int x^2 \sqrt[3]{x^3+3}\ dx&=\int x^2 \sqrt[3]{t}\ \dfrac{dt}{3x^2}\\
&=\frac13\int t^{\large\frac13}\ dt\\
&=\frac13\cdot\frac1{\frac13+1}t^{\large\frac13+1}+C\\
&=\frac14t^{\large\frac43}+C\\
&=\frac14(x^3+3)^{\large\frac43}+C.
\end{align}
$$
A: Take the differential form ($dx$) out from underneath the squareroot sign. It reads $\int x^2\cdot (x^3+3)^{1/3} dx = 1/3 \int (3x^2 dx)\cdot (x^3+3)^{1/3} = 1/3\cdot (x^3+3)^{4/3} / (4/3) + c = \frac{(x^3+3)^{4/3}}{4} + c$.
A: Make the substitution $t=x^3+3$. Then $dt=3x^2dx$, and we have
$$
\frac{1}{3}\int t^{1/3}dt = \frac{1}{3}\left(\frac{3}{4}t^{4/3}\right) + C=\frac{1}{4}\left( x^3+3\right)^{4/3} +C.
$$
A: Notice that $3x^2$ is the derivative of $x^3+2$. 
So if you set $t=x^3+2$, you get $\dfrac{dx}{dt}=3x^2$ which gives $dx = 3x^2 dt$.
Now the whole thing becomes:
$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{3}\int 3x^2\sqrt[3]{x^3+3} \ dx = \frac13 \int t^{1/3} dt = \frac{1}{3}\frac{1}{1+\frac{1}{3}} t^{1+\frac{1}{3}} + C$$
$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{4}(x^2+3)^{4/3} + C$$
A: If $dt=3x^2\,dx$ then $\dfrac{dt}{3} = x^2\,dx$.
