Question

The value of $\int \limits \frac {(x^2-1)dx}{x^3 \sqrt {2x^4-2x^2+1}}$ is

• A. $2 \sqrt {2- \frac {2}{x^2}+ \frac {1}{x^4}}+c$
• B. $2 \sqrt {2+ \frac {2}{x^2}+ \frac {1}{x^4}}+c$
• C. $\frac {1}{2} \sqrt {2- \frac {2}{x^2}+ \frac {1}{x^4}}+c$
• D. None of these

Answer 1

Answer: (C)

$\frac {1}{2} \sqrt {2- \frac {2}{x^2}+ \frac {1}{x^4}}+c$

Answer 2

Let I$= \int \limits \frac {(x^2-1)dx}{x^3 \sqrt {2x^4-2x^2+1}}$ \hspace10mm [dividing \, numerator \, and \, enominator \, by \, x^5] \hspace15mm = \int \limits \frac { \bigg (\frac {1}{x^3} - \frac {1}{x^5} \bigg )dx }{\sqrt {2- \frac {2}{x^2} + \frac {1}{x^4}}} Put$2- \frac {2}{x^2}+ \frac {1}{x^4}=t \Rightarrow \bigg ( \frac {4}{x^3}- \frac {4}{x^5} \bigg ) dx=dt$ \therefore \hspace10mm I= \frac {1}{4} \int \limits \frac {dt}{ \sqrt t}= \frac {1}{4}. \frac {t^{1/2}}{1/2}+c \hspace15mm = \frac {1}{2} \sqrt {2- \frac {2}{x^2}+ \frac {1}{x^4}}+c