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Q.
If $\int \frac{x^2-1}{x^3 \sqrt{3 x^4+2 x^2-1}} d x=f(x)+C$, where $f(1)=-1$, then $|f(-1)|$ is equal to
Integrals
Solution:
$\int \frac{x^{-3}-x^{-5}}{\sqrt{3+2 x^{-2}-x^{-4}}} d x$
$3+2 x^{-2}-x^{-4}=t^2 $
$\left(-4 x^{-3}+4 x^{-5}\right) d x=2 t d t$
$\left( x ^{-3}- x ^{-5}\right) dx =-\frac{1}{2} tdt$
$\therefore I=\int \frac{-\frac{1}{2} t d t}{t}=-\frac{1}{2} t+C=-\frac{1}{2} \sqrt{3+2 x^{-2}-x^{-4}}+C$
$f(x)=-\frac{1}{2} \sqrt{3+2 x^{-2}-x^{-4}}$
$f(-1)=-\frac{1}{2} \sqrt{3+2-1}=-1 $
$\Rightarrow | f (-1)|=1$