Understanding substitution to compute primitive

Question

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?

Answer 1

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 $$ \begin{align} \int x^2 \sqrt[3]{x^3+3}\ dx&=\int x^2 \sqrt[3]{t}\ \dfrac{dt}{3x^2}\\ &=\frac13\int t^{\large\frac13}\ dt\\ &=\frac13\cdot\frac1{\frac13+1}t^{\large\frac13+1}+C\\ &=\frac14t^{\large\frac43}+C\\ &=\frac14(x^3+3)^{\large\frac43}+C. \end{align} $$

Answer 2

Take the differential form ($dx$) out from underneath the squareroot sign. It reads $\int x^2\cdot (x^3+3)^{1/3} dx = 1/3 \int (3x^2 dx)\cdot (x^3+3)^{1/3} = 1/3\cdot (x^3+3)^{4/3} / (4/3) + c = \frac{(x^3+3)^{4/3}}{4} + c$.

Answer 3

Make the substitution $t=x^3+3$. Then $dt=3x^2dx$, and we have $$ \frac{1}{3}\int t^{1/3}dt = \frac{1}{3}\left(\frac{3}{4}t^{4/3}\right) + C=\frac{1}{4}\left( x^3+3\right)^{4/3} +C. $$

Answer 4

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:

$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{3}\int 3x^2\sqrt[3]{x^3+3} \ dx = \frac13 \int t^{1/3} dt = \frac{1}{3}\frac{1}{1+\frac{1}{3}} t^{1+\frac{1}{3}} + C$$

$$\int x^2 \sqrt[3]{x^3+3}\ dx = \frac{1}{4}(x^2+3)^{4/3} + C$$

Answer 5

If $dt=3x^2\,dx$ then $\dfrac{dt}{3} = x^2\,dx$.