Integral Calculus Question 151
Question: If $ \int{x\log ( 1+\frac{1}{x} )dx} $ $ =f(x)\log (x+1)+g(x)x^{2}+Lx+C $ , then
Options:
A) $ f(x)=\frac{1}{2}x^{2} $
B) $ g(x)=\log x $
C) $ L=1 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \int{x\log ( 1+\frac{1}{x} )dx} $ $ =\log ( 1+\frac{1}{x} ).\frac{x^{2}}{2}-\int{\frac{x}{x+1}\cdot ( -\frac{1}{x^{2}} )\cdot \frac{x^{2}}{2}dx} $ $ =\frac{x^{2}}{2}\log ( \frac{x+1}{x} )\cdot \frac{x^{2}}{2}+\frac{1}{2}\int{\frac{x+1-1}{x+1}}dx $ $ =\frac{x^{2}}{2}\log ( \frac{x+1}{x} )+\frac{1}{2}x-\frac{1}{2}\log (x+1)+c $ $ =( \frac{x^{2}-1}{2} )\log (x+1)-\frac{x^{2}}{2}\log x+\frac{1}{2}x+c $
BETA
