Integral Calculus Question 53
Question: To find the value of $ \int_{{}}^{{}}{\frac{1+\log x}{x}}dx $ , the proper substitution is
[MP PET 1988]
Options:
A) $ \log x=t $
B) $ 1+\log x=t $
C) $ \frac{1}{x}=t $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ 1+\log x=t\Rightarrow \frac{1}{x},dx=dt $ Therefore, $ \int_{{}}^{{}}{\frac{1+\log x}{x}},dx=\int_{{}}^{{}}{t,dt}=\frac{t^{2}}{2} $ $ =\frac{{{(1+\log x)}^{2}}}{2} $ .
BETA
