Question

$\int\frac {dx}{x\sqrt{x^6-16}} $= is equal to :

• A. $Sec^{-1}(\frac{x^3}{4})+c $
• B. $\frac {1}{12} Sec^{-1}(\frac{x^3}{4})+c $
• C. $Cosh^{-1}(\frac{x^3}{4})+c $
• D. $\frac {1}{3} Sec^{-1}(\frac{x^3}{4})+c $

Answer 1

Answer: (B)

Let $ I=\int \frac{d x}{x \sqrt{x^{6}-16}} $
$=\frac{1}{3} \int \frac{3 x^{2}}{x^{3} \sqrt{\left(x^{3}\right)^{2}-4^{2}}} d x$
Put $x^{3} =t $
$\Rightarrow 3 x^{2} d x =d t $
$\therefore I =\frac{1}{3} \int \frac{d t}{t \sqrt{t^{2}-4^{2}}}$
$=\frac{1}{3 \times 4} \sec ^{-1}\left(\frac{t}{4}\right)+c $
$=\frac{1}{12} \sec ^{-1}\left(\frac{x^{3}}{4}\right)+c$