Integral Calculus Question 92
Question: $ \int {{{ \frac{\log x-1}{1+{{(log,x)}^{2}}} }}^{2}} $ dx is equal to
Options:
A) $ \frac{\log ,x}{{{(log,x)}^{2}}+1}+c $
B) $ \frac{x}{x^{2}+1}+c $
C) $ \frac{xe^{x}}{1+x^{2}}+c $
D) $ \frac{x}{{{(log,x)}^{2}}+1}+c $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \int{\frac{{{(\log x-1)}^{2}}}{{{(1+{{(\log x)}^{2}})}^{2}}}dx} $ $ =\int{[ \frac{1}{(1+{{(\log x)}^{2}})}-\frac{2\log x}{{{(1+{{(\log x)}^{2}})}^{2}}} ]}dx $ $ =\int{[ \frac{e^{t}}{1+t^{2}}-\frac{2te^{t}}{{{(1+t^{2})}^{2}}} ]}dt $ (Putting $ \log x=t\Rightarrow dx=e^{t}dt $ ) $ \int{e^{t}[ \frac{1}{1+t^{2}}-\frac{2t}{{{(1+t^{2})}^{2}}} ]dt} $ $ =\frac{e^{t}}{1+t^{2}}+c $ $ =\frac{x}{1+{{(\log x)}^{2}}}+c $ (Resubstituting $ t=\log x $ )
BETA
