1708\hspace{-16}$\;\bf{(2)}$\;\;\; $Given $\bf{\int\frac{\left(x^2-1\right)}{x^3\sqrt{2x^4-2x^2+1}}dx}$\\\\\\ We can write it as $\bf{\int\frac{(x^2-1)}{x^3\cdot x^2\sqrt{2-2x^{-2}+x^{-4}}}dx}$\\\\\\ We can Write it as $\bf{\int\frac{(x^{-3}-x^{-5})}{\sqrt{2-2x^{-2}+x^{-4}}}dx}$\\\\\\ Now Let $\bf{\left(2-2x^{-2}+x^{-4}\right)=t^2\;,}$ Then $\bf{\left(x^{-3}-x^{-5}\right)dx = \frac{t}{2}dt}$\\\\\\ So Integral is $\bf{=\frac{1}{2}\int\frac{t}{t}dt = \frac{1}{2}\sqrt{2-2x^{-2}+x^{-4}}+\mathbb{C}}$
- man111 singh For (1) Multiply Numerator and Denemonitor by (x+1)Upvote·0· Reply ·2014-07-21 00:01:43
Sushovan Halder thanks to both of you.
- Sushovan Halder in the first integral idid the same and took y=x+1/x but after that the integral is becoming too hard for me to evaluate
