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

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

Grade:12th pass

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
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

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