<s>[INST] 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?
 [/INST] 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:
$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{3}\int 3x^2\sqrt[3]{x^3+3} \ dx = \frac13 \int t^{1/3} dt = \frac{1}{3}\frac{1}{1+\frac{1}{3}} t^{1+\frac{1}{3}} + C$$
$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{4}(x^2+3)^{4/3} + C$$
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