Question

If $\int \sqrt{x^2-14}dx=Ax\sqrt{x^2-14}+B\ln |x+\sqrt{x^2-14}|+C$ (where $A,B$ are fixed constants and $C$ is integration constant). Then the value of $\left|\frac{B}{A}\right|$ is

Answer 1

We know,
$\int \sqrt{x^2-a^2}=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln |x+\sqrt{x^2-a^2}|+C$
Let $I=\int \sqrt{x^2-14}dx$
$\Rightarrow I=\int \sqrt{x^2-(\sqrt{14}{)}^2}dx\Rightarrow I=\frac{x}{2}\sqrt{x^2-14}-\left(\frac{14}{2}\right)\ln |x+\sqrt{x^2-14}|+C\Rightarrow A=\frac{1}{2};B=-7\therefore \left|\frac{B}{A}\right|=14$