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) \[-\frac{\sqrt{{{x}^{2}}-1}}{x}+c\]
Correct Answer: A
Solution :
Put \[x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta \,d\theta ,\] then \[\int_{{}}^{{}}{\frac{1}{{{x}^{2}}\sqrt{1+{{x}^{2}}}}\,dx=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\theta \,d\theta }{{{\tan }^{2}}\theta \sec \theta }=\int_{{}}^{{}}{\text{cosec}\,\theta \,\text{cot}\theta \,d\theta }}}\] \[=-\text{cosec}\,\theta +\text{c}=\frac{\text{--}\sqrt{{{x}^{\text{2}}}+1}}{x}+c.\]
You need to login to perform this action.
You will be redirected in
3 sec
