Integral Calculus Question 371
Question: $ \int_{{}}^{{}}{e^{x}\frac{(x^{2}+1)}{{{(x+1)}^{2}}}dx=} $
Options:
A) $ ( \frac{x-1}{x+1} )e^{x}+c $
B) $ e^{x}( \frac{x+1}{x-1} )+c $
C) $ e^{x}(x+1)(x-1)+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{e^{x}(x^{2}+1)}{{{(x+1)}^{2}}},dx}=\int_{{}}^{{}}{\frac{e^{x}(x^{2}-1+2)}{{{(x+1)}^{2}}},dx} $ $ =\int_{{}}^{{}}{e^{x}[ \frac{x-1}{x+1}+\frac{2}{{{(x+1)}^{2}}} ]},dx=\int_{{}}^{{}}{e^{x}[f(x)+{f}’(x)],dx} $ where $ f(x)=\frac{x-1}{x+1} $ and $ {f}’(x)=\frac{2}{{{(x+1)}^{2}}}=e^{x}( \frac{x-1}{x+1} )+c $ .
BETA
