Integral Calculus Question 217
Question: What is $ \int{\frac{\log x}{{{(1+\log ,x)}^{2}}}dx} $ equal to?
Options:
A) $ \frac{1}{{{( 1+\log x )}^{3}}}+c $
B) $ \frac{1}{{{( 1+\log x )}^{2}}}+c $
C) $ \frac{x}{( 1+\log x )}+c $
D) $ \frac{x}{{{( 1+\log x )}^{2}}}+c $ Where c is a constant.
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Answer:
Correct Answer: C
Solution:
[c] Let $ I=\int{\frac{\log x}{{{(1+\log x)}^{2}}}dx} $ Put $ \log x=t\Rightarrow \frac{1}{x}dx=dt $ $ I=\int{\frac{e^{t}t}{{{(1+t)}^{2}}}dt}=\int{\frac{e^{t}.(t+1-1)}{{{(1+t)}^{2}}}dt} $ $ =\int{\frac{e^{t}(1+t)}{{{(1+t)}^{2}}}dt-\int{\frac{e^{t}}{{{(1+t)}^{2}}}dt}} $ $ =\int{\frac{e^{t}}{1+t}dt-\int{\frac{e^{t}}{{{(1+t)}^{2}}}dt}} $ $ =\frac{e^{t}}{1+t}-\int{-e^{t}\frac{1}{{{(1+t)}^{2}}}dt-\int{\frac{e^{t}}{{{(1+t)}^{2}}}dt}} $ $ =\frac{e^{t}}{1+t}+\int{^{e^{t}}\frac{1}{{{(1+t)}^{2}}}dt-\int{\frac{e^{t}}{{{(1+t)}^{2}}}dt}} $ $ =\frac{e^{t}}{1+t}+c=\frac{x}{1+\log x}+c $
BETA
