Integral Calculus Question 31
Question: $ \int_{{}}^{{}}{\frac{1}{x-x^{3}}\ dx=} $
[MP PET 1996]
Options:
A) $\frac{1}{2} \log | \frac{1-x^2}{x^2}|+c$
B) $\log |\frac{1-x}{x(1+x)}| +c$
C) $\log | x(1-x^2) |+c$
D) $\frac{1}{2} \log |\frac{x^2}{1-x^2}|+c$
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{1}{x-x^{3}} dx=\int_{{}}^{{}}{\frac{1}{x(1+x)(1-x)} dx}} $ $ =\frac{1}{2}\int_{{}}^{{}}{( \frac{2}{x}-\frac{1}{1+x}+\frac{1}{1-x} ) dx} $ $ =\frac{1}{2}[2\log x-\log (1+x)-\log (1-x)]=\frac{1}{2}\log \frac{x^{2}}{(1-x^{2})}+c $ .
BETA
