Question

$int(4x^{3}+9)sqrt{x^{4}+9x}dx$

Explanation:

Step1: Use substitution

Let $u = x^{4}+9x$. Then $du=(4x^{3} + 9)dx$.

Step2: Rewrite the integral

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

Step3: Integrate $\sqrt{u}$

We know that $\sqrt{u}=u^{\frac{1}{2}}$, and $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n
eq - 1$). So $\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{2}{3}u^{\frac{3}{2}}+C$.

Step4: Substitute back $u$

Substitute $u = x^{4}+9x$ back into the result. We get $\frac{2}{3}(x^{4}+9x)^{\frac{3}{2}}+C$.

Answer:

$\frac{2}{3}(x^{4}+9x)^{\frac{3}{2}}+C$