Please help me with $\int 3x^2\sqrt{5x^3-3}dx$

Question

So, according to photomath the solution is:

however all I can do is:2:

Answer 1

Here is an evaluation with explained substitution:

$$\int 3x^2\sqrt{5x^3-3}dx$$

$u=5x^3-3\implies du=5\times 3x^2-0\implies \frac15 du=3x^2 dx$. We clearly see this part in the integral. Notice the (reverse) power rule:

$$\int \sqrt{5x^3-3}\boxed{3x^2dx}=\frac15 \int \sqrt u du=\frac15 \int u^\frac 12du=C+\frac15 \frac{1}{\frac12+1}u^{\frac 12+1}=\frac{2u^\frac32}{15}+C$$

Let’s substitute back $u=5x^3-3$: $$\int 3x^2\sqrt{5x^3-3}dx =\frac{2u^\frac32}{15}+C= C+\frac{2(5x^3-3)^\frac32}{15}= C+\frac{2\sqrt{5x^3-3}(5x^3-3)}{15} $$

I hope this helps. Please remember to add more detail and use MathJax. Please correct me and give me feedback!