Question

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

Explanation:

Step1: Use substitution

Let $u = x^{2}+2x$. Then $du=(2x + 2)dx=2(x + 1)dx$. First, complete the square for $u$: $u=x^{2}+2x=(x + 1)^{2}-1$. And we can rewrite the integral $\int\sqrt{x^{2}+2x}dx=\int\sqrt{(x + 1)^{2}-1}dx$. Let $t=x + 1$, then $x=t - 1$ and $dx=dt$, and the integral becomes $\int\sqrt{t^{2}-1}dt$.

Step2: Use integral formula

The integral formula for $\int\sqrt{t^{2}-a^{2}}dt=\frac{t}{2}\sqrt{t^{2}-a^{2}}-\frac{a^{2}}{2}\ln|t+\sqrt{t^{2}-a^{2}}|+C$ (here $a = 1$). So $\int\sqrt{t^{2}-1}dt=\frac{t}{2}\sqrt{t^{2}-1}-\frac{1}{2}\ln|t+\sqrt{t^{2}-1}|+C$.

Step3: Substitute back

Since $t=x + 1$, we have $\int\sqrt{x^{2}+2x}dx=\frac{x + 1}{2}\sqrt{x^{2}+2x}-\frac{1}{2}\ln|x + 1+\sqrt{x^{2}+2x}|+C$.

Answer:

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