Integral Calculus Question 42
Question: $ \int_{{}}^{{}}{\frac{x}{x^{4}-1}dx=} $
Options:
A) $ \frac{1}{4}\log [ \frac{x^{2}-1}{x^{2}+1} ]+c $
B) $ \frac{1}{4}\log [ \frac{x^{2}+1}{x^{2}-1} ]+c $
C) $ \frac{1}{2}\log [ \frac{x^{2}-1}{x^{2}+1} ]+c $
D) $ \frac{1}{2}\log [ \frac{x^{2}+1}{x^{2}-1} ]+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{x}{(x^{4}-1)},dx=\frac{1}{2}\int_{{}}^{{}}{[ \frac{x}{x^{2}-1}-\frac{x}{x^{2}+1} ]},dx} $ $ =\frac{1}{4}\log (x^{2}-1)-\frac{1}{4}\log (x^{2}+1)=\frac{1}{4}\log ( \frac{x^{2}-1}{x^{2}+1} )+c $ . Aliter : Put $ t=x^{2}\Rightarrow dt=2x,dx, $ then $ \int_{{}}^{{}}{\frac{x}{x^{4}-1},dx}=\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{t^{2}-1}=\frac{1}{2}.\frac{1}{2}\log \frac{t-1}{t+1}+c} $ $ =\frac{1}{4}\log [ \frac{x^{2}-1}{x^{2}+1} ]+c. $
BETA
