Integral Calculus Question 450
Question: $ \int_{{}}^{{}}{\frac{x^{2}+1}{x(x^{2}-1)}\ dx} $ is equal to
[MP PET 1999]
Options:
A) $ \log \frac{x^{2}-1}{x}+c $
B) $ -\log \frac{x^{2}-1}{x}+c $
C) $ \log \frac{x}{x^{2}+1}+c $
D) $ -\log \frac{x}{x^{2}+1}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{\frac{x^{2}+1}{x(x^{2}-1)},dx}=\int_{{}}^{{}}{\frac{1+( \frac{1}{x^{2}} )}{x-( \frac{1}{x} )},dx} $ Put $ x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{x^{2}}),dx=dt $
$ \therefore ,I=\int_{{}}^{{}}{\frac{dt}{t}}=\log t+c=\log \frac{x^{2}-1}{x}+c $ .
BETA
