Question

Find the Integral
$\sf \int\dfrac{x^{2} - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}}dx$
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Answer 1

Question :

Integrate :

$\sf\int\dfrac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$

Solution :

We have to integrate

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

$\sf\:I=\int\dfrac{x^2-1}{x^3\sqrt{x^4(2-\frac{2}{x^2}+\frac{1}{x^4}})}dx$

$\sf=\int\dfrac{x^2-1}{x^3\times\:x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$

$\sf=\int\dfrac{x^2-1}{x^5\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$

$\sf=\int(\dfrac{x^2-1}{x^5})\times\dfrac{1}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$

$\sf=\int(\dfrac{x^2}{x^5}-\dfrac{1}{x^5})\times\dfrac{1}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx...(1)$

Let $\sf\:t=2-\dfrac{2}{x^2}+\dfrac{1}{x^4}$

Now Differentiate with respect to x

$\sf\dfrac{dt}{dx}=\dfrac{d(2)}{dx}-\dfrac{2x{}^{-2}}{dx}+\dfrac{x{}^{-4}}{dx}$

$\sf\dfrac{dt}{dx}=0+\dfrac{4}{x^3}-\dfrac{4}{x^5}$

$\sf\dfrac{dt}{dx}=4(\dfrac{1}{x^3}-\dfrac{1}{x^5})$

$\sf\dfrac{dt}{4}=(\dfrac{1}{x^3}-\dfrac{1}{x^5})dx$

Then,

$\sf\int(\dfrac{x^2}{x^5}-\dfrac{1}{x^5})\times\dfrac{1}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}dx$

$\sf=\int\dfrac{1}{\sqrt{t}}\times\dfrac{dt}{4}$

$\sf=\dfrac{1}{4}\int\:t{}^{\frac{-1}{2}}dt$

We know that $\sf\int\:x^n=\dfrac{x{}^{n+1}}{n+1}$

$\sf=\dfrac{1}{4}\times\dfrac{\sqrt{t}}{\frac{1}{2}}+c$

$\sf=\dfrac{2}{4}\times\sqrt{t}+c$

$\sf=\dfrac{\sqrt{t}}{2}+c$

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

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

$\sf=\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+c$