Question

If $\int \frac{x^2-1}{x^3\sqrt{3x^4+2x^2-1}}dx=f(x)+C$, where $f(1)=-1$, then $|f(-1)|$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

$\int \frac{x^{-3}-x^{-5}}{\sqrt{3+2x^{-2}-x^{-4}}}dx$
$3+2x^{-2}-x^{-4}=t^2$
$\left(-4x^{-3}+4x^{-5}\right)dx=2tdt$
$\left(x^{-3}-x^{-5}\right)dx=-\frac{1}{2}tdt$
$\therefore I=\int \frac{-\frac{1}{2}tdt}{t}=-\frac{1}{2}t+C=-\frac{1}{2}\sqrt{3+2x^{-2}-x^{-4}}+C$
$f(x)=-\frac{1}{2}\sqrt{3+2x^{-2}-x^{-4}}$
$f(-1)=-\frac{1}{2}\sqrt{3+2-1}=-1$
$\Rightarrow |f(-1)|=1$