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Integral Calculus> Integration of [(x^2 -1)/(x^3 (sqrt(2x^4 ...

question mark

Integration of [(x^2 -1)/(x^3 (sqrt(2x^4 -2x^2 +1)))]dx=?

John Doe , 11 Years ago

Grade 12th pass

1 Answers

Jitender Singh

Ans:

$I = \int \frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}}dx$

$u = x^{2}$

$du = 2x.dx$

$I = \frac{1}{2}\int \frac{\frac{1}{u}-\frac{1}{u^{2}}}{\sqrt{2u^{2}-2u+1}}dx$

$I = \frac{1}{2}\int \frac{\frac{1}{u}-\frac{1}{u^{2}}}{\sqrt{(\sqrt{2}u-\frac{1}{\sqrt{2}})^{2}+\frac{1}{2}}}dx$

$s = \sqrt{2}u-\frac{1}{\sqrt{2}}$

$ds = \sqrt{2}du$

$I = \frac{1}{2\sqrt{2}}\int \frac{\frac{2\sqrt{2}}{2s+\sqrt{2}}-\frac{8}{(2s+\sqrt{2})^{2}}}{\sqrt{s^{2}+\frac{1}{2}}}ds$

$s = \frac{tan(t)}{\sqrt{2}}$

$ds = \frac{sec^{2}(t)}{\sqrt{2}}dt$

$I = \frac{1}{2\sqrt{2}}\int (\frac{2\sqrt{2}}{\sqrt{2}tan(t)+\sqrt{2}}-\frac{8}{(\sqrt{2}tan(t)+\sqrt{2})^{2}})dt$

$w = tan(\frac{t}{2})$

$dw = \frac{1}{2}.sec^{2}(\frac{t}{2})dt$

$I = \frac{1}{2\sqrt{2}}\int \frac{4(w^{2}+2w-1)}{w^{4}-4w^{3}+2w^{2}+4w+1}dw$

Just apply the partial fraction here, you will get

$I = \frac{\sqrt{2x^{4}-2x^{2}+1}}{2x^{2}} + constant$

Thanks & Regards

Jitender Singh

IIT Delhi

askIITians Faculty

Last Activity: 11 Years ago

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