Question

Evaluate: $\int \frac{6x+7}{\sqrt{(4-x)(5-x)}}dx(x<4)$.

Answer 1

$\int \frac{6x+7}{\sqrt{(4-x)(5-x)}}dx$
$=\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx$
$=\int \frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}dx$
$=3\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx+34\int \frac{1}{\sqrt{x^2-9x+20}}dx$ -(1)
$I_1$ $I_2$
$I_1=3\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx$
Let $t=x^2-9x+20$
Differentiating w.r.t to x:
$dt=(2x-9)dx$
on substituting:
$I_1=3\int \frac{1}{\sqrt{t}}dt$
$=3\int t^{-\frac{1}{2}}dt$
$=6\sqrt{t}$ + $C_1$ ${[\int t^{n}dt=\frac{t^{n+1}}{n+1}+c]}_{n\ne 1}$
on Resubstituting:
$I_1=6\sqrt{x^2-9x+20}$ + $C_1$
$I_2=34\int \frac{1}{\sqrt{x^2-9x+20}}dx$
$=34\int \frac{1}{\sqrt{x^2-9x+20}}dx$
$=34\int \frac{1}{\sqrt{x^2-9x+\frac{81}{4}-\frac{81}{4}+20}}dx$
$=34\int \frac{1}{\sqrt{{(x-\frac{9}{2})}^2-\frac{1}{4}}}dx$
Let $u=x-\frac{9}{2}$
Differentiating w.r.t to x:
$du=dx$
on substituting:
$=34\int \frac{1}{\sqrt{u^2-\frac{1}{4}}}dx$
$=34\ln [u+\sqrt{u^2-\frac{1}{4}}]$ + $C_2$
On Resubstituting:
$=34\ln [x-\frac{9}{2}+\sqrt{{(x-\frac{9}{2})}^2-\frac{1}{4}}]$ + $C_2$


$I_2=34\ln [x-\frac{9}{2}+\sqrt{x^2-9x+20}]$ + $C_2$ $\left[\int \frac{1}{\sqrt{u^2-a^2}}dx=\ln [u+\sqrt{u^2-a^2}]+C\right]$
From $e{q}^n\left(1\right):$
$\int \frac{6x+7}{\sqrt{(4-x)(5-x)}}dx$ = $6\sqrt{x^2-9x+20}$ + $C_1$ + $34\ln [x-\frac{9}{2}+\sqrt{x^2-9x+20}]$ + $C_2$
$\int \frac{6x+7}{\sqrt{(4-x)(5-x)}}dx$ = $6\sqrt{x^2-9x+20}$ + $34\ln [x-\frac{9}{2}+\sqrt{x^2-9x+20}]$ + $C$ $[C=C_1+C_2]$