Integral Calculus Question 272
Question: $ \int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{2}}}\ dx=} $
Options:
A) $ \log (x+1)+\frac{2}{x+1}+c $
B) $ \log (x+1)-\frac{2}{x+1}+c $
C) $ \frac{2}{x+1}-\log (x+1)+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{2}}},dx=\int_{{}}^{{}}{\frac{x+1-2}{{{(x+1)}^{2}}}},dx} $ $ =\int_{{}}^{{}}{\frac{1}{x+1},dx}-\int_{{}}^{{}}{\frac{2}{{{(x+1)}^{2}}},dx=\log (x+1)+\frac{2}{(x+1)}+c} $ .
BETA
