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Evaluate the integral from 0 to 2 of 375x5(x2+1)-4

by RoRi

Show that

    \[ \int_0^2 375x^5 (x^2+1)^{-4} \, dx = 2^n \]

for some n \in \mathbb{Z}.

Proof. We evaluate the integral and show that it is a power of 2. (Possibly Apostol has a clever idea in mind to show this a power of 2 without directly evaluating the integral, but I don’t see it.)

Make the substitution,

    \[ u = x^2 + 1 \qquad \implies \qquad du = 2x \, dx. \]

This gives us new bounds of integration

    \[ u(0) = 1 \qquad \text{and} \qquad u(2) = 5. \]

And thus,

    \begin{align*} \int_0^2 375x^5 (x^2+1)^{-4} \, dx &= \frac{375}{2} \int_0^2 x^4 (x^2+1)^{-4} 2x \, dx \\ &= \frac{375}{2} \int_1^5 (u-1)^2 u^{-4} \, du \\ &= \frac{375}{2} \int_1^5 \frac{u^2 - 2u + 1}{u^4} \, du \\ &= \frac{375}{2} \left( \int_1^5 \frac{1}{u^2} \, du - 2\int_1^5 \frac{1}{u^3} \, du + \int_1^5 \frac{1}{u^4} \, du \right) \\[9pt] &= \frac{375}{2}\left. \left( -\frac{1}{u} + \frac{1}{u^2} - \frac{1}{3 u^3} \right) \right|_1^5 \\ &= \frac{375}{2} \left( -\frac{1}{5} + \frac{1}{25} - \frac{1}{375} + 1 - 1 + \frac{1}{3} \right) \\ &= \frac{375}{2} \left( -\frac{61}{375} + \frac{1}{3} \right) \\ &= -\frac{61}{2} + \frac{375}{6} \\ &= \frac{375}{6} - \frac{183}{6} \\ &= \frac{192}{6} \\ &= 32 \\ &= 2^5. \qquad \blacksquare \end{align*}

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