Question

find the indefinite integral.
int \frac { x } { sqrt { x - 4 } } d x
int \frac { x } { sqrt { x - 4 } } d x =

Explanation:

Step1: Use substitution

Let \(u = \sqrt{x - 4}\), then \(x=u^{2}+4\) and \(dx = 2u\ du\).
The integral \(\int\frac{x}{\sqrt{x - 4}}dx\) becomes \(\int\frac{u^{2}+4}{u}\cdot2u\ du=\int2(u^{2}+4)du\).

Step2: Integrate term - by - term

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have:
\(\int2(u^{2}+4)du=2\int u^{2}du+8\int du\).
\(2\times\frac{u^{3}}{3}+8u+C=\frac{2}{3}u^{3}+8u + C\).

Step3: Substitute back

Since \(u=\sqrt{x - 4}\), the result is \(\frac{2}{3}(x - 4)^{\frac{3}{2}}+8\sqrt{x - 4}+C\).

Answer:

\(\frac{2}{3}(x - 4)^{\frac{3}{2}}+8\sqrt{x - 4}+C\)