problem 5: $int_{1}^{5}\frac{x}{sqrt{2x - 1}}dx$

Explanation:

Step1: Use substitution

Let $u = 2x - 1$, then $x=\frac{u + 1}{2}$ and $dx=\frac{1}{2}du$. When $x = 1$, $u=2\times1 - 1=1$; when $x = 5$, $u=2\times5 - 1 = 9$.

Step2: Rewrite the integral

The integral $\int_{1}^{5}\frac{x}{\sqrt{2x - 1}}dx$ becomes $\int_{1}^{9}\frac{\frac{u + 1}{2}}{\sqrt{u}}\times\frac{1}{2}du=\frac{1}{4}\int_{1}^{9}\frac{u + 1}{\sqrt{u}}du=\frac{1}{4}\int_{1}^{9}(u^{\frac{1}{2}}+u^{-\frac{1}{2}})du$.

Step3: Integrate term - by - term

We know that $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $\frac{1}{4}\int_{1}^{9}(u^{\frac{1}{2}}+u^{-\frac{1}{2}})du=\frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}}+2u^{\frac{1}{2}}\right]_{1}^{9}$.

Step4: Evaluate the definite integral

$\frac{1}{4}\left[\left(\frac{2}{3}\times9^{\frac{3}{2}}+2\times9^{\frac{1}{2}}\right)-\left(\frac{2}{3}\times1^{\frac{3}{2}}+2\times1^{\frac{1}{2}}\right)\right]=\frac{1}{4}\left[\left(\frac{2}{3}\times27 + 2\times3\right)-\left(\frac{2}{3}+2\right)\right]=\frac{1}{4}\left[(18 + 6)-\left(\frac{2 + 6}{3}\right)\right]=\frac{1}{4}\left[24-\frac{8}{3}\right]=\frac{1}{4}\times\frac{72 - 8}{3}=\frac{1}{4}\times\frac{64}{3}=\frac{16}{3}$.

Answer:

$\frac{16}{3}$