Integral Calculus Question 434
Question: The value of $ \int_{{}}^{{}}{\frac{dx}{x\sqrt{x^{4}-1}}} $ is
Options:
A) $ \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 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 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 $
BETA
