Question

If $\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3}dx=\ln |x|+\frac{a}{x^b}+C$, then absolute value of $\left(\frac{b}{a}\right)$ is (where $a,b$ are fixed contants and $C$ is constant of integration)

Answer 1

$\sqrt{(x^4+x^{-4}+2)}=\sqrt{{(x^2+\frac{1}{x^2})}^2}$
$\therefore \int \frac{\sqrt{x^4+x^{-4}+2}}{x^3}dx=\int \left(\frac{x^2+\frac{1}{x^2}}{x^3}\right)dx$
$\Rightarrow \int (\frac{1}{x}+\frac{1}{x^5})dx$
$=\ln |x|+\frac{x^{-4}}{-4}+C$
$=\ln |x|-\frac{1}{4x^4}+C$
$\Rightarrow a=-\frac{1}{4},b=4$
$\Rightarrow$ absolute value of $\frac{b}{a}=16$