$int\frac{(4x - 1)dx}{(2x + 1)^{\frac{3}{2}}}$

Explanation:

Step1: Use substitution

Let $u = 2x+1$, then $x=\frac{u - 1}{2}$ and $dx=\frac{1}{2}du$. Also, $4x-1=4\times\frac{u - 1}{2}-1=2u-2 - 1=2u-3$. The integral becomes $\int\frac{2u - 3}{u^{\frac{3}{2}}}\times\frac{1}{2}du=\frac{1}{2}\int(\frac{2u}{u^{\frac{3}{2}}}-\frac{3}{u^{\frac{3}{2}}})du=\int(u^{-\frac{1}{2}}-\frac{3}{2}u^{-\frac{3}{2}})du$.

Step2: Integrate term - by - term

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have: $\int u^{-\frac{1}{2}}du=2u^{\frac{1}{2}}+C_1$ and $\int\frac{3}{2}u^{-\frac{3}{2}}du=- 3u^{-\frac{1}{2}}+C_2$. So, $\int(u^{-\frac{1}{2}}-\frac{3}{2}u^{-\frac{3}{2}})du=2u^{\frac{1}{2}}+3u^{-\frac{1}{2}}+C$.

Step3: Substitute back $u = 2x + 1$

We get $2(2x + 1)^{\frac{1}{2}}+3(2x + 1)^{-\frac{1}{2}}+C$.

Answer:

$2(2x + 1)^{\frac{1}{2}}+3(2x + 1)^{-\frac{1}{2}}+C$