Question

Integrate the following functions.
$\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}dx$

Answer 1

$\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}dx=\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx$
Let $6x+7=A\frac{d}{dx}(x^2-9x+20)+B$
$\Rightarrow 6x+7=A(2x-9)+B\Rightarrow 6x+7=2Ax+(-9A+B)$
On equating the coefficients of x and constant term on both sides, we get
$2A=6\Rightarrow A=3and-9A+B=7\Rightarrow B=34$
$\Rightarrow \int \frac{6x+7}{\sqrt{x^2-9x+20}}dx=\int \frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}dx=3\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx+34\int \frac{1}{\sqrt{x^2-9x+20}}dx$

Let $I_1=\int \frac{2x-9}{\sqrt{x^2-9x+20}}dxand{I}_2=\int \frac{1}{\sqrt{x^2-9x+20}}dx$
$\therefore \int \frac{6x+7}{\sqrt{x^2-9x+20}}dx=3I_1+34I_2......\left(i\right)$
Now, $I_1=\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx$
Let $x^2-9x+20=t\Rightarrow (2x-9)dx=dt$
$\therefore I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}+C_1=2\sqrt{x^2-9x+20}+C_1....\left(ii\right)$
And $I_2=\int \frac{1}{\sqrt{x^2-9x+20}}dx$
$x^2-9x+20\text{can be written as}{x}^2-9x+20+\frac{81}{4}-\frac{81}{4}$
Therefore, $x^2-9x+20+\frac{81}{4}-\frac{81}{4}={(x-\frac{9}{2})}^2-\frac{1}{4}={(x-\frac{9}{2})}^2-(\frac{1}{2}{)}^2$
$\therefore I_2=\int \frac{1}{\sqrt{(x-\frac{9}{2}{)}^2}-(\frac{1}{2}{)}^2}dx=log\left|\right(x-\frac{9}{2})+\sqrt{(x-\frac{9}{2}{)}^2-\frac{1}{4}}|+C_2$

$[\because \int \frac{dx}{\sqrt{x^2-a^2}}=log|x+\sqrt{x^2-a^2}|]=log\left|\right(x-\frac{9}{2})+\sqrt{x^2-9x+20}|+C_2......\left(iii\right)$
On substituting the values of $I_1$ and $I_2$ from Eqs.(ii)and (iii)in Eq. (i). we get
$\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx=3\left[2\sqrt{x^2-9x+20}\right]+34log\left|\right(x-\frac{9}{2})+\sqrt{x^2-9x+20}|+C[\because 3C_1+34C_2=C]=6\sqrt{x^2-9x+20}+34log\left|\right(x-\frac{9}{2})+\sqrt{x^2-9x+20}|+C$