Integral Calculus Question 179
Question: $ \int_{{}}^{{}}{\frac{(x+1){{(x+\log x)}^{2}}}{x}dx=} $
[AI CBSE 1986]
Options:
A) $ \frac{1}{3}(x+\log x)+c $
B) $ \frac{1}{3}{{(x+\log x)}^{2}}+c $
C) $ \frac{1}{3}{{(x+\log x)}^{3}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ t=x+\log x\Rightarrow dt=( 1+\frac{1}{x} ),dx, $ then $ \int_{{}}^{{}}{\frac{(x+1){{(x+\log x)}^{2}}}{x},dx}=\int_{{}}^{{}}{t^{2}dt}=\frac{t^{3}}{3}+c $ $ =\frac{1}{3}{{(x+\log x)}^{3}}+c. $
BETA
