Integral Calculus Question 87
Question: $ \int \frac{\sqrt{x-1}}{x\sqrt{x+1}}dx $ is equal to
Options:
A) $ \ln ,| x-\sqrt{x^{2}-1} |-{{\tan }^{-1}}x+c $
B) $ \ln ,| x+\sqrt{x^{2}-1} |-{{\tan }^{-1}}x+c $
C) $ \ln ,| x-\sqrt{x^{2}-1} |-{{\sec }^{-1}}x+c $
D) $ \ln ,| x+\sqrt{x^{2}-1} |-{{\sec }^{-1}}x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ I=\int{\frac{\sqrt{x-1}}{x\sqrt{x+1}}dx} $ $ =\int{\frac{x-1}{x\sqrt{x^{2}-1}}dx} $ $ =\int{\frac{dx}{\sqrt{x^{2}-1}}-\int{\frac{dx}{x\sqrt{x^{2}-1}}}} $ $ =\ln |x+\sqrt{x^{2}-1}|-{{\sec }^{-1}}x+c $
BETA
