Integral Calculus Question 207
Question: $ \int_{{}}^{{}}{\frac{{{(1+\log x)}^{2}}}{x}}\ dx= $
[Roorkee 1977]
Options:
A) $ {{(1+\log x)}^{3}}+c $
B) $ 3{{(1+\log x)}^{3}}+c $
C) $ \frac{1}{3}{{(1+\log x)}^{3}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ (1+\log x)=t\Rightarrow \frac{1}{x}dx=dt $ $ \int_{{}}^{{}}{\frac{{{(1+\log x)}^{2}}}{x},dx=\int_{{}}^{{}}{t^{2}dt}} $ $ =\frac{t^{2}}{3}+c=\frac{{{(1+\log x)}^{3}}}{3}+c. $
BETA
