Question

make a substitution to express the integrand as a rational function and then evaluate the integral. (remember the constant of integration.)
int\frac{x^{3}}{sqrt{x^{2}+9}}dx
check your answer by taking the derivative.

Explanation:

Step1: Make a substitution

Let $u = x^{2}+9$, then $du = 2x dx$ and $x^{2}=u - 9$. Also, $x dx=\frac{1}{2}du$. The integral $\int\frac{x^{3}}{\sqrt{x^{2}+9}}dx=\int\frac{x^{2}\cdot x}{\sqrt{x^{2}+9}}dx$. Substituting $u$ and $x dx$ values, we get $\frac{1}{2}\int\frac{u - 9}{\sqrt{u}}du$.

Step2: Rewrite the integrand

$\frac{1}{2}\int\frac{u - 9}{\sqrt{u}}du=\frac{1}{2}\int(\frac{u}{\sqrt{u}}-\frac{9}{\sqrt{u}})du=\frac{1}{2}\int(u^{\frac{1}{2}}-9u^{-\frac{1}{2}})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\frac{1}{2}(\frac{u^{\frac{3}{2}}}{\frac{3}{2}}-9\frac{u^{\frac{1}{2}}}{\frac{1}{2}})+C=\frac{1}{3}u^{\frac{3}{2}}-9u^{\frac{1}{2}}+C$.

Step4: Substitute back $u = x^{2}+9$

The result is $\frac{1}{3}(x^{2}+9)^{\frac{3}{2}}-9(x^{2}+9)^{\frac{1}{2}}+C$.

Answer:

$\frac{1}{3}(x^{2}+9)^{\frac{3}{2}}-9(x^{2}+9)^{\frac{1}{2}}+C$