1 x x^4-1 d x=? - Vidyadip

MCQ

$\int \frac{1}{x \sqrt{x^4-1}} d x=?$

A
$\operatorname{cosec}^{-1} x^2+C$
✓
$\frac{1}{2} \sec ^{-1} x^2+C$
C
$\sec ^{-1} x^2+C$
D
$2 \operatorname{cosec}^{-1} x^2+C$

✓

Answer

Correct option: B.

$\frac{1}{2} \sec ^{-1} x^2+C$

Formula $:- \int x^n d x=\frac{x^{n+1}}{n+1}+c ; \int \frac{1}{t \sqrt{t^2-1}} d t=\sec ^{-1} t+c$
Therefore,
Put $x^2=t$
$\Rightarrow 2 xdx=dt$
$=\int \frac{1}{x \sqrt{t^2-1}} \times \frac{d t}{2 x} \Rightarrow \frac{1}{2} \int \frac{1}{t \sqrt{t^2-1}} d t$
$=\frac{1}{2} \sec ^{-1} t+c$
$=\frac{1}{2} \sec ^{-1} x^2+c$

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