Definite Integration Question 408
Question: $ \int_0^{1}{\frac{e^{x}(x-1)}{{{(x+1)}^{3}}}dx=} $
[SCRA 1986]
Options:
A) $ \frac{e}{4} $
B) $ \frac{e}{4}-1 $
C) $ \frac{e}{4}+1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_0^{1}{\frac{e^{x}(x-1)}{{{(x+1)}^{3}}}}dx=\int_0^{1}{\frac{e^{x}(x+1-2)}{{{(x+1)}^{3}}}}dx $
$ \int_0^{1}{\frac{e^{x}}{{{(x+1)}^{2}}}}dx-2\int_0^{1}{\frac{e^{x}}{{{(x+1)}^{3}}}}dx=[ \frac{e^{x}}{{{(x+1)}^{2}}} ]_0^{1}=\frac{e}{4}-1 $ .
BETA
