Integral Calculus Question 3
Question: $ {{\int{{ \frac{(\log x-1)}{1+{{(\log x)}^{2}}} }}}^{2}}dx $ is equal to
[AIEEE 2005]
Options:
A) $ \frac{xe^{x}}{1+x^{2}}+c $
B) $ \frac{x}{{{(\log x)}^{2}}+1}+C $
C) $ \frac{\log x}{{{(\log x)}^{2}}+1}+c $
D) $ \frac{x}{x^{2}+1}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\int{{ \frac{\log x-1}{1+{{(\log x)}^{2}}} }}}^{2}}dx $ . Put $ \log x=t\Rightarrow dx=e^{t}dt $ \ Integral $ =\int{e^{t}[ \frac{1}{1+t^{2}}-\frac{2t}{{{(1+t^{2})}^{2}}} ]}\ dt $ $ [ \because \ \int{e^{x}[f(x)+f’(x)]\ dx=e^{x}f(x)+c}\ ] $ $ =\frac{e^{t}}{1+t^{2}}+C=\frac{x}{1+{{(\log x)}^{2}}}+C $ .
BETA
