Question

evaluate the integral. (remember to use absolute values where appropriate. use c for the constant of integration.)
\int\frac{x^{2}+2x}{\sqrt{x^{2}+2x}}dx

Explanation:

Step1: Use substitution

Let $u = x^{2}+2$, then $du = 2x dx$.

Step2: Rewrite the integral

The integral $\int\sqrt{x^{2}+2x}dx$ becomes $\int\sqrt{u}\frac{du}{2}=\frac{1}{2}\int u^{\frac{1}{2}}du$.

Step3: Apply power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). So, $\frac{1}{2}\int u^{\frac{1}{2}}du=\frac{1}{2}\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$.

Step4: Simplify the expression

$\frac{1}{2}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}u^{\frac{3}{2}}+C$.

Step5: Substitute back $u = x^{2}+2$

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

Answer:

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