23-11-09: Evaluate the integral

∂∂∂∂∂∂∂∂
∞∞∞∞∞
∫∫∫∫∫∫∫∫

lol...wat sort of an answer is tat organic[3]

Okay let me modify the problem... Just prove that the given integral is a rational number.

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This is an ugly problem, yet quite instructive. The limits of the integrations are not at all random.
Write the given expression as I₁ + I₂ + I₃.
Let
f(x)=\dfrac{x^2-x}{x^3-3x+1}
Apply the substitution t = 11-x to I₃. The limit of the integration now becomes -100 and -10. Further, x = 1 - 1t, so dx = 1t² dt
Note that this substitution leaves f unchanged. i.e.
f\left(1-\dfrac{1}{t}\right)=f(t)
Hence, we get
I_3=\int_{-100}^{-10}\dfrac{f(t)^2}{t^2}\ \mathrm dt=\int_{-100}^{-10}\dfrac{f(x)^2}{x^2}\ \mathrm dx
Next apply the substitution z = 1 &ndash 1x to I₂. i.e. x = 11-z. This gives
\mathrm dx = \dfrac{1}{(1-z)^2}\ \mathrm dz
And the limits become -100 and -10. Again note that
f\left(\dfrac{1}{1-z}\right)=f(z)
So I_3=\int_{-100}^{-10}\dfrac{f(z)^2}{(1-z)^2}\ \mathrm dz
According, the given expression is
I=\int_{-100}^{-10}f(x)^2\left(1+\dfrac{1}{(1-x)^2}+\dfrac{1}{x^2}\right)\ \mathrm dx
which on simplification becomes
I=\int_{-100}^{-10}\dfrac{x^4-2x^3+3x^2-2x+1}{(x^3-3x+1)^2}\ \mathrm dx
It does not seems to simplified. However, the final "trick". Let's look at f'(x).
f'(x)=\dfrac{(x^3-3x+1)(2x-1)-(x^2-x)(3x^2-3)}{(x^3-3x+1)^2}
Simplify the above expression gives
f'(x)=-\dfrac{x^4-2x^3+3x^2-2x+1}{(x^3-3x+1)^2}
which is precisely the integrand!!. Hence,
I=-\int_{-100}^{-10}f'(x)\ \mathrm dx=f(-100)-f(-10)=\dfrac{11131110}{107634259}

Of course you could use that? But the question is --- "Will that help?"

yes .SIR .it is of no help!!!

thank u sir !!!!!!!!!!!!!!!

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