Question

If $\int \frac{6x+7}{\sqrt{\left(x-5\right)\left(x-4\right)}}dx$
$=A\sqrt{x^2-9x+20}+Blog\left|x+\sqrt{x^2-9x+20}-\frac{9}{2}\right|+C$
Then the values of A and B are

• A. $6,34$
• B. $3,9$
• C. $12,17$
• D. None of these

Answer 1

Answer: (A)

Let $6x+7=\lambda \frac{d}{dx}\left(x-5\right)\left(x-4\right)+\mu$
i.e. $6x+7=\lambda \left(2x-9\right)+\mu$ which gives $\lambda =3$ and $\mu =34$
$\therefore \int \frac{6x+7}{\sqrt{\left(x-5\right)\left(x-4\right)}}dx=\int \frac{3\left(2x-9\right)+34}{x^2-9x+20}dx$
$=3\int \left(2x-9\right){\left(x^2-9x+20\right)}^{-\frac{1}{2}}dx+34\int \frac{dx}{\sqrt{x^2-9x+20}}$
$=3\int \frac{{\left(x^2-9x+20\right)}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+34\int \frac{dx}{\sqrt{x^2-9x+\frac{81}{4}-\frac{1}{4}}}$
$=6\sqrt{x^2-9x+20}+34\int \frac{dx}{\sqrt{{\left(x-\frac{9}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}}$
$=6\sqrt{x^2-9x+20}+34log\left\{\left|x-\frac{9}{2}+\sqrt{{\left(x-\frac{9}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}\right|\right\}+C$
$=6\sqrt{x^2-9x+20}+34log\left|x+\sqrt{x^2-9x+20}-\frac{9}{2}\right|+C$
$\therefore A=6,B=34.$