Integral Calculus Question 268
Question: $ \int_{{}}^{{}}{\frac{1}{x^{2}\sqrt{1+x^{2}}}}\ dx= $
Options:
A) $ -\frac{\sqrt{1+x^{2}}}{x}+c $
B) $ \frac{\sqrt{1+x^{2}}}{x}+c $
C) $ -\frac{\sqrt{1-x^{2}}}{x}+c $
D) $ -\frac{\sqrt{x^{2}-1}}{x}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta ,d\theta , $ then $ \int_{{}}^{{}}{\frac{1}{x^{2}\sqrt{1+x^{2}}},dx=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\theta ,d\theta }{{{\tan }^{2}}\theta \sec \theta }=\int_{{}}^{{}}{cosec,\theta ,cot\theta ,d\theta }}} $ $ =-cosec,\theta +c=\frac{\text{–}\sqrt{{x^{2}}+1}}{x}+c. $
BETA
