Question
$\int \frac{(2 x-7)}{\sqrt{4 x-1}} d x$

Answer

Let $I =\int \frac{2 x-7}{\sqrt{4 x-1}} d x$
Let $2 x -7= A (4 x -1)+ B$
$\therefore 2 x-7=4 A x+(B-A)$
By equating the coefficients on both sides, we get
$4 A=2$ and $B-A=-7$
$\therefore A=\frac{1}{2}$ and $B=-7+A$
$=-7+\frac{1}{2}$
$=-\frac{13}{2}$
$\therefore 2 x-7=\frac{1}{2}(4 x-1)-\frac{13}{2}$
$\therefore I=\frac{1}{2} \int \frac{(4 x-1)-13}{\sqrt{4 x-1}} d x$
$=\frac{1}{2} \int\left(\frac{4 x-1}{\sqrt{4 x-1}}-\frac{13}{\sqrt{4 x-1}}\right) d x$
$=\frac{1}{2} \int\left(\sqrt{4 x-1}-\frac{13}{\sqrt{4 x-1}}\right) d x$
$=\frac{1}{2} \int(4 x-1)^{\frac{1}{2}} d x-\frac{13}{2} \int(4 x-1)^{\frac{1}{2}} d x$
$=\frac{1}{2}\left[\frac{(4 x-1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{4}\right]-\frac{13}{2}\left[\frac{(4 x-)^{\frac{1}{2}}}{\frac{1}{2}} \times \frac{1}{4}\right]$
$\therefore I =\frac{1}{12}(4 x-)^{\frac{3}{2}}-\frac{13}{4} \sqrt{4 x-1}+ c$

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