Integral Calculus Question 337
Question: $ \int_{{}}^{{}}{\frac{\log x}{{{(1+\log x)}^{2}}}dx=} $
Options:
A) $ \frac{1}{1+\log x}+c $
B) $ \frac{x}{{{(1+\log x)}^{2}}}+c $
C) $ \frac{x}{1+\log x}+c $
D) $ \frac{1}{{{(1+\log x)}^{2}}}+c $
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Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{\log x}{{{(1+\log x)}^{2}}},dx} $ . Put $ 1+\log x=t\Rightarrow \frac{1}{x}dx=dt $
$ \Rightarrow dx=x,dt={e^{t-1}}dt, $ then it reduces to $ \int_{{}}^{{}}{\frac{(t-1),{e^{t-1}}}{t^{2}}dt}=\int_{{}}^{{}}{{e^{t-1}}( \frac{1}{t}-\frac{1}{t^{2}} ),dt}=\frac{{e^{t-1}}}{t}=\frac{x}{1+\log x}+c $ .
BETA
