$\\int 4x^{3}(x^{4}+4)^{9}dx = \\square$

Explanation:

Step1: Use substitution

Let $u = x^{4}+4$, then $du=4x^{3}dx$.

Step2: Rewrite the integral

The integral $\int 4x^{3}(x^{4} + 4)^{9}dx$ becomes $\int u^{9}du$.

Step3: Integrate $u^{9}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int u^{9}du=\frac{u^{10}}{10}+C$.

Step4: Substitute back $u$

Substitute $u = x^{4}+4$ back into the result, we get $\frac{(x^{4}+4)^{10}}{10}+C$.

Answer:

$\frac{(x^{4}+4)^{10}}{10}+C$