 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/7a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

With

u = x^{4}

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{1}{4} \int_{0}^{1} u^{- 3/4} (1 - u)^{- 1/2} d u = \frac{1}{4} B (\frac{1}{4}, \frac{1}{2}) .

Γ (1/2) = π

\frac{1}{4} B (\frac{1}{4}, \frac{1}{2}) = \frac{Γ ( 1/4 ) π}{4Γ ( 3/4 )}, Γ (1/4) Γ (3/4) = π 2 .

Therefore

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{Γ ( 1/4 ) ^{2}}{32 π} .

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