Question
Evaluate:
$\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x$

Answer

$ \int \frac{2 x-7}{\sqrt{4 x-1}} d x$
$= \frac{1}{2} \int \frac{2(2 x-7)}{\sqrt{4 x-1}} d x$
$=  \frac{1}{2} \int \frac{(4 x-1)-13}{\sqrt{4 x-1}} d x$
$=\frac{1}{2} \int\left(\frac{4 x-1}{\sqrt{4 x-1}}-\frac{13}{\sqrt{4 x-1}}\right) d x$
$=\frac{1}{2} \int(4 x-1)^{\frac{1}{2}} d x-\frac{13}{2} \int(4 x-1)^{-\frac{1}{2}} d x$
$=\frac{1}{2} \cdot \frac{(4 x-1)^{\frac{3}{2}}}{(4)\left(\frac{3}{2}\right)}-\frac{13}{2} \cdot \frac{(4 x-1)^{\frac{1}{2}}}{(4)\left(\frac{1}{2}\right)}+c$
$=\frac{1}{12}(4 x-1)^{\frac{3}{2}}-\frac{13}{4} \sqrt{4 x-1}+c$.

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