Question

what is the function $u$?
answer: $u = 2x^{2}$
what is $du$?
answer: $du=xdx$
check your variables - you might be using an incorrect one.
$dx$
find the value(s) of $a$ and $b$ (if appropriate), and use the formula to evaluate the integral. use the letter c for an arbitrary constant.
answer: $int\frac{2x^{3}dx}{sqrt{2 + 4x^{2}}}=\frac{2}{3}(x^{2}-1)sqrt{2 + 4x^{2}}+c$

Explanation:

Step1: Define substitution

Let $u = 2 + 4x^{2}$, then $du=8xdx$.

Step2: Rewrite the integrand

We have $\int\frac{2x^{3}dx}{\sqrt{2 + 4x^{2}}}=\frac{1}{4}\int\frac{x^{2}\cdot8xdx}{\sqrt{2 + 4x^{2}}}$. Since $u = 2+4x^{2}$, then $x^{2}=\frac{u - 2}{4}$.

Step3: Substitute and integrate

The integral becomes $\frac{1}{16}\int\frac{(u - 2)du}{\sqrt{u}}=\frac{1}{16}\int(u^{\frac{1}{2}}-2u^{-\frac{1}{2}})du$.
Integrating term - by - term: $\frac{1}{16}(\frac{2}{3}u^{\frac{3}{2}}-4u^{\frac{1}{2}})+C$.

Step4: Substitute back

Substitute $u = 2 + 4x^{2}$ back into the result: $\frac{1}{24}(2 + 4x^{2})^{\frac{3}{2}}-\frac{1}{4}(2 + 4x^{2})^{\frac{1}{2}}+C=\frac{1}{12}(x^{2}+ \frac{1}{2})(2 + 4x^{2})^{\frac{1}{2}}-\frac{1}{4}(2 + 4x^{2})^{\frac{1}{2}}+C=\frac{1}{6}(x^{2}-1)\sqrt{2 + 4x^{2}}+C$.

Answer:

$\int\frac{2x^{3}dx}{\sqrt{2 + 4x^{2}}}=\frac{1}{6}(x^{2}-1)\sqrt{2 + 4x^{2}}+C$