Integral Calculus Question 197
Question: If $ A=\int\limits_0^{1}{\frac{e^{t}}{t+1}dt,} $ then $ \int\limits_0^{1}{e^{t}\log (1+t)dt} $ in terms of A equals
Options:
A) $ e\log (A) $
B) $ \frac{e}{2}-A $
C) $ e-l-\frac{A}{2} $
D) $ \frac{e}{2}-l-A $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \int\limits_0^{1}{e^{t}}\log (1+t)dt=[ e^{t}.\frac{1}{1+t} ]_0^{1}-\int\limits_0^{1}{\frac{e^{t}}{1+t}dt} $ $ =\frac{e}{2}-1-A $
BETA
