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Finding value of $$\frac{8}{\pi}\int^{2}_{0}\frac{1}{x^2-x+1+\sqrt{x^4-4x^3+9x^2-10x+5}}$$

Let $$I=\int^{2}_{0}\frac{1}{x^2-x+1+\sqrt{x^4-4x^3+9x^2-10x+5}}dx$$

$$I=\int^{2}_{0}\frac{1}{x(x-1)+\sqrt{(x^2-2x)^2+5x(x-2)+5)}}dx$$

From $\displaystyle \int^{a}_{0}f(x)dx=\int^{a}_{0}f(a-x)dx$

$$I=\int^{2}_{0}\frac{1}{(2-x)(1-x)+\sqrt{(2-x)^2x^2+5(2-x)(-x)+5}}dx$$

How do i solve it help me please

jacky
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    Any particular reason for computing this integral? Would an accurate numerical approximation do the trick? If you make the change of variable $y=x-1$ you are able to eliminate some terms inside the square root and get $$I = \int_{-1}^1 \dfrac{dy}{1+y+y^2+\sqrt{y^4+3y^2+1}}$$ – PierreCarre May 28 '19 at 08:45
  • @PierreCarre This is the key. See my answer. – popi May 28 '19 at 08:49

1 Answers1

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Let's start multiplying the original fraction (up and down) by $$x^2-x+1-\sqrt{x^4-4x^3+9x^2-10x+5}$$

to obtain

$$\frac{1}{2}\int^{2}_{0}\frac{x^2-x+1-\sqrt{x^4-4x^3+9x^2-10x+5}}{x^3- 3 x^2+4 x -2}dx$$

Now, taking acount:

  • $x^2-x+1=(x-1)^2+(x-1)+1$
  • $x^4-4x^3+9x^2-10x+5=(x-1)^4+3(x-1)^2+1$
  • $ x^3- 3 x^2+4 x -2 =(x-1)((x-1)^2+1)$

we perfom the change $x-1=t$ to transform our integral as follow

$$\frac{1}{2}\int^{1}_{-1}\frac{t^2+t+1-\sqrt{t^4+3t^2+1}}{t(t^2+1)}dt=\frac{1}{2}\int_{-1}^1\frac{t}{t^2+1}+\frac{1}{t^2+1}+\frac{1}{t(t^2+1)}-\frac{\sqrt{t^4+3t^2+1}}{t(t^2+1)}dt$$

As first, third and fourth fraction are odd function over symmetric interval, they vanishes. Only rest the second fraction and then our integral reads

$$\int^{2}_{0}\frac{1}{x^2-x+1+\sqrt{x^4-4x^3+9x^2-10x+5}}dx=\frac{1}{2}\int_{-1}^1\frac{1}{t^2+1}dt=\frac{\pi}{4}$$

Thus

Finding value of $$\frac{8}{\pi}\int^{2}_{0}\frac{1}{x^2-x+1+\sqrt{x^4-4x^3+9x^2-10x+5}}=2$$

popi
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