Discussion
\(\int x^{7}(x^{8} + 1)^{-3/4}\,dx =\)
Answer: E
The integral is \(\frac{1}{2}(x^{8} + 1)^{1/4} + C\).
Put \(u = x^{8} + 1\), so \(du = 8x^{7}\,dx\):
\[ \begin{aligned} \int x^{7}(x^{8} + 1)^{-3/4}dx &= \frac{1}{8}\int u^{-3/4}du \\ &= \frac{1}{8}\cdot 4u^{1/4} + C = \frac{1}{2}u^{1/4} + C \end{aligned} \]
\(4(x^{8} + 1)^{1/4}\) forgets the factor \(\frac{1}{8}\), and \((x^{8} + 1)^{1/4}\) applies it only partly. The forms with \(1 + \frac{1}{x^{8}}\) belong to a different integrand, \(x^{-9}(1 + x^{-8})^{-3/4}\).
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