19∂∂∂∂∂∂∂∂
∞∞∞∞∞
∫∫∫∫∫∫∫∫
Evaluate the following integral:
\int_{-100}^{-10}\left(\dfrac{x^2-x}{x^3-3x+1}\right)^2\mathrm dx+\int_{\frac{1}{101}}^{\frac{1}{11}}\left(\dfrac{x^2-x}{x^3-3x+1}\right)^2\mathrm dx + \int_{\frac{101}{100}}^{\frac{11}{10}}\left(\dfrac{x^2-x}{x^3-3x+1}\right)^2\mathrm dx
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11 Answers
66Okay let me modify the problem... Just prove that the given integral is a rational number.
1sir i am getting very very weird answers!!!!!
ans is like dis SIR ji
a typo sir
each integral is squared
how to go furder sir
66This 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}
66Of course you could use that? But the question is --- "Will that help?"