Question

evaluate the integral. (remember to use absolute values where appropriate. use c for the constant of integration.)
∫√(x² + 2x) dx

Explanation:

Step1: Complete the square

First, complete the square for the expression inside the square - root: $x^{2}+2x=(x + 1)^{2}-1$. So the integral becomes $\int\sqrt{(x + 1)^{2}-1}dx$. Let $u=x + 1$, then $du=dx$ and the integral is $\int\sqrt{u^{2}-1}du$.

Step2: Use the integral formula for $\sqrt{u^{2}-a^{2}}$

The integral formula for $\int\sqrt{u^{2}-a^{2}}du=\frac{u}{2}\sqrt{u^{2}-a^{2}}-\frac{a^{2}}{2}\ln|u+\sqrt{u^{2}-a^{2}}|+C$ (where $a = 1$ in our case).
Substituting $a = 1$ into the formula, we get $\int\sqrt{u^{2}-1}du=\frac{u}{2}\sqrt{u^{2}-1}-\frac{1}{2}\ln|u+\sqrt{u^{2}-1}|+C$.

Step3: Substitute back $u=x + 1$

Substituting $u=x + 1$ back into the result, we have $\frac{x + 1}{2}\sqrt{(x + 1)^{2}-1}-\frac{1}{2}\ln|(x + 1)+\sqrt{(x + 1)^{2}-1}|+C=\frac{x + 1}{2}\sqrt{x^{2}+2x}-\frac{1}{2}\ln|x + 1+\sqrt{x^{2}+2x}|+C$.

Answer:

$\frac{x + 1}{2}\sqrt{x^{2}+2x}-\frac{1}{2}\ln|x + 1+\sqrt{x^{2}+2x}|+C$