A) \[\frac{\log x}{x}+\frac{1}{{{x}^{2}}}+C\]
B) \[-\frac{\log x}{x}+\frac{2}{x}+C\]
C) \[-\frac{\log x}{x}-\frac{1}{2x}+C\]
D) \[x\log x+\frac{1}{{{x}^{2}}}+C\]
E) \[-\frac{\log x}{x}-\frac{1}{x}+C\]
Correct Answer: E
Solution :
\[\int{\frac{\log x}{{{x}^{2}}}}dx\] Put \[t=\log x\] \[dt=\frac{1}{x}.dx\] \[\Rightarrow \] \[x\,dt=dx\] \[\Rightarrow \] \[\int{\frac{t.x\,dt}{{{x}^{2}}}}=\int{\frac{t}{{{e}^{t}}}}dt\] \[\int{\underset{I}{\mathop{t}}\,\underset{II}{\mathop{{{e}^{-t}}}}\,dt}=[t(-{{e}^{-t}})-\int{1.(-{{e}^{-t}})}dt]\] \[=-t{{e}^{-t}}-{{e}^{-t}}+C\] \[=-\frac{t}{{{e}^{t}}}-\frac{1}{{{e}^{t}}}+C\] \[=-\frac{\log x}{x}-\frac{1}{x}+C\]
You need to login to perform this action.
You will be redirected in
3 sec
