Integrate the function 6 x+7 (x-5)(x-4) - Vidyadip

Question

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

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Answer

Let 6x + 7 = $A \frac{d}{d x}\left(x^{2}-9 x+20\right)+B$
$\Rightarrow$ 6x + 7 = A(2x - 9) + B
Now, equating the coefficients of x and constant term on both sides, we get,
2A = 6
$\Rightarrow$ A = 3
Also, -9A + B = 7
$\Rightarrow$ B = 34
$\Rightarrow$ 6x + 7 = 3 (2x - 9) + 34
$\therefore~ \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x$
$=\int \frac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x$
$= 3 \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x+34 \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$
Now, in $\int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$
Let x² - 9x + 20 = t
$\Rightarrow$ (2x - 9)dx = dt
$\therefore \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x=\int \frac{d t}{\sqrt{t}}$
$=2 \sqrt{t}$
$=2 \sqrt{x^{2}-9 x+20}$ ...(i)
And in $\int \frac{1}{\sqrt{x^{2}-9 x+20}} d x$, we have
x² - 9x + 20 = $x^{2}-9 x+20+\frac{81}{4}-\frac{81}{4}$
$= \left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$
$\Rightarrow \int \frac{1}{\sqrt{\mathrm{x}^{2}-9 \mathrm{x}+20}} \mathrm{dx}=\int \frac{1}{\sqrt{\left(\mathrm{x}-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} \mathrm{d} \mathrm{x}$
= $\log \left| (x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20} |$ .....(ii)
Thus, from (i) and (ii), we get,
$ \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x=3[2 \sqrt{x^{2}-9 x+20}]+34\left[\log \left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9 x+20}\right|\right]+C$
$= 6 \sqrt{\mathrm{x}^{2}-9 \mathrm{x}+20}+34 \log \left[\left(\mathrm{x}-\frac{9}{2}\right)+\sqrt{\mathrm{x}^{2}-9 \mathrm{x}+20}\right]+\mathrm{C}$

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