Question

Integrate the function $\frac{6x+7}{\sqrt{(x-5)(x-4)}}$

Answer 1

$\frac{6x+7}{\sqrt{(x-5)(x-4)}}=\frac{6x+7}{\sqrt{x^2-9x+20}}$
Let $6x+7=A\frac{d}{dx}(x^2-9x+20)+B$
$\Rightarrow 6x+7=A(2x-9)+B$
Equating the coefficients of $x$ and constant term, we obtain
$2A=6\Rightarrow A=3$
& $-9A+B=7\Rightarrow B=34$
$\therefore 6x+7=3(2x-9)+34$
$\Rightarrow \int \frac{6x+7}{\sqrt{x^2-9x+20}}=\int \frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}dx$
$=3\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx+34\int \frac{1}{\sqrt{x^2-9x+20}}dx$
Let $I_1=\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx$ and $I_2=\frac{1}{\sqrt{x^2-9x+20}}dx$
$\therefore \int \frac{6x+7}{\sqrt{x^2-9x+20}}=3I_1+34I_2$ ..........(1)
Then,
$I_1=\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx$
Let $x^2-9x+20=t$
$\Rightarrow (2x-9)dx=dt$
$\Rightarrow I_1=\frac{dt}{\sqrt{t}}$
$I_1=2\sqrt{t}$
$I_1=2\sqrt{x^2-9x+20}$ ......(2)
and $I_2=\int \frac{1}{\sqrt{x^2-9x+20}}dx$
$x^2-9x+20$ can be written as $x^2-9x+20+\frac{81}{4}-\frac{81}{4}$
Therefore,
$x^2-9x+20+\frac{81}{4}-\frac{81}{4}$
$={(x-\frac{9}{2})}^2-\frac{1}{4}$
$={(x-\frac{9}{2})}^2-{\left(\frac{1}{2}\right)}^2$
$\Rightarrow I_2=\int \frac{1}{\sqrt{{(x-\frac{9}{2})}^2-{\left(\frac{1}{2}\right)}^2}}dx$
$\Rightarrow I_2=\log |(x-\frac{9}{2})+\sqrt{x^2-9x+20}|$ ........(3)
Substituting equations (2) and (3) in (1), we obtain
$\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx=3\left[2\sqrt{x^{2}9x+20}\right]+34\log [(x-\frac{9}{2})+\sqrt{x^2-9x+20}]+C$
$=6\sqrt{x^2-9x+20}+34\log \left[2\sqrt{x^{2}9x+20}\right]+34\log [(x-\frac{9}{2})+\sqrt{x^2-9x+20}]+C$