find the indefinite integral. (remember the constant of integration.)\n int\frac{x + 5}{sqrt{4-(x - 2)^2}}dx

Explanation:

Step1: Split the integral

We split $\int\frac{x + 5}{\sqrt{4-(x - 2)^2}}dx$ into $\int\frac{x}{\sqrt{4-(x - 2)^2}}dx+\int\frac{5}{\sqrt{4-(x - 2)^2}}dx$.

Step2: First - integral substitution

Let $u=x - 2$, then $x=u + 2$ and $dx=du$. The first integral $\int\frac{x}{\sqrt{4-(x - 2)^2}}dx=\int\frac{u + 2}{\sqrt{4 - u^2}}du=\int\frac{u}{\sqrt{4 - u^2}}du+\int\frac{2}{\sqrt{4 - u^2}}du$. For $\int\frac{u}{\sqrt{4 - u^2}}du$, let $t = 4 - u^2$, then $dt=-2udu$ and $\int\frac{u}{\sqrt{4 - u^2}}du=-\frac{1}{2}\int\frac{dt}{\sqrt{t}}=-\sqrt{t}=-\sqrt{4 - u^2}=-\sqrt{4-(x - 2)^2}$. For $\int\frac{2}{\sqrt{4 - u^2}}du$, since $\int\frac{1}{\sqrt{a^2 - x^2}}dx=\arcsin(\frac{x}{a})+C$ ($a = 2$ here), $\int\frac{2}{\sqrt{4 - u^2}}du=2\arcsin(\frac{u}{2})=2\arcsin(\frac{x - 2}{2})$.

Step3: Second - integral

The second integral $\int\frac{5}{\sqrt{4-(x - 2)^2}}dx$. Since $\int\frac{1}{\sqrt{a^2 - x^2}}dx=\arcsin(\frac{x}{a})+C$ ($a = 2$), $\int\frac{5}{\sqrt{4-(x - 2)^2}}dx=5\arcsin(\frac{x - 2}{2})$.

Step4: Combine results

Combining the results of the two - part split, we have $-\sqrt{4-(x - 2)^2}+2\arcsin(\frac{x - 2}{2})+5\arcsin(\frac{x - 2}{2})+C=-\sqrt{4-(x - 2)^2}+7\arcsin(\frac{x - 2}{2})+C$.

Answer:

$-\sqrt{4-(x - 2)^2}+7\arcsin(\frac{x - 2}{2})+C$