The value of _ ^ dx x x^4 - 1 is - Vidyadip

MCQ

The value of $\int_{}^{} {\frac{{dx}}{{x\sqrt {{x^4} - 1} }}} $ is

✓
$\frac{1}{2}{\sec ^{ - 1}}{x^2} + k$
B
$\log x\sqrt {{x^4} - 1} + k$
C
$x\log \sqrt {{x^4} - 1} + k$
D
$\log \sqrt {{x^4} - 1} + k$

✓

Answer

Correct option: A.

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

a
(a) $I = \int_{}^{} {\frac{{dx}}{{x\sqrt {{x^4} - 1} }}} $
Put ${x^2} = t \Rightarrow 2x\,dx = dt \Rightarrow dx = \frac{{dt}}{{2x}} = \frac{{dt}}{{2\sqrt t }}$
$\therefore \,\,\,I = \int_{}^{} {\frac{{dt}}{{2t\sqrt {{t^2} - 1} }}} = \frac{1}{2}{\sec ^{ - 1}}t + k = \frac{1}{2}{\sec ^{ - 1}}{x^2} + k$

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