Question

Evaluate:
$\int \frac{1}{(2x-7)}\sqrt{(x-3)(x-4)}dx$

Answer 1

$I=\int \frac{dx}{(2x-7)}\sqrt{(x-3)(x-4)}$

$=\int \frac{\sqrt{(x-3)(x-4)}}{(2x-7)}dx$

$=\int \frac{\sqrt{x^2-3x-4x+12}}{2x-7}dx$

$=\int \frac{\sqrt{x^2-7x+12}}{2x-7}dx$

$=\int \frac{\sqrt{x^2-2.\frac{7}{2}x+{\left(\frac{7}{2}\right)}^2-{\left(\frac{7}{2}\right)}^2+12}}{2x-7}dx$

$=\int \frac{\sqrt{{(x-\frac{7}{2})}^2+12-\frac{49}{4}}}{2x-7}dx$

$=\int \frac{\sqrt{{(x-\frac{7}{2})}^2-{\left(\frac{1}{2}\right)}^2}}{2(1-\frac{7}{2})}dx$

Let ${(x-\frac{7}{2})}^2-{\left(\frac{1}{2}\right)}^2=t^2$

$\Rightarrow 2(x-\frac{7}{2})dx=2tdt$ $\Rightarrow dx=\frac{2tdt}{2(x-\frac{7}{2})}$

$\Rightarrow I=\int \frac{t.tdt}{2{(x-\frac{7}{2})}^2}$

$=\frac{1}{4}\int \frac{t^{2}dt}{{(x-\frac{7}{2})}^2}$

$=\frac{1}{4}\int \frac{t^2+\frac{1}{4}-\frac{1}{4}dt}{t^2+\frac{1}{4}}$

$=\frac{1}{4}\int 1dt-\frac{1}{4}\int \frac{1}{t^2+\frac{1}{4}}dt$

$=\frac{1}{4}t-\frac{1}{4\times \frac{1}{4}}{\tan }^{-1}\frac{t}{\frac{1}{4}}+c$

$=\frac{1}{4}t-{\tan }^{-1}\frac{t}{\frac{1}{4}}+c$

$=\frac{1}{4}\sqrt{x^2-7x+12}-{\tan }^{-1}4\sqrt{(x-3)(x-4)}+c$

Hence, the answer is $\frac{1}{4}\sqrt{x^2-7x+12}-{\tan }^{-1}4\sqrt{(x-3)(x-4)}+c.$