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Calculus IChapter 5: IntegralsSection 5-3

Substitution Rule for Indefinite Integrals — Question 10

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Question 10

Evaluate the integral ∫xln⁡(1−x2)dx.\int x \ln(1-x^2)\,dx.

Original worksheet page 1: question and worked solution for 5-3-010
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Question 10 - Solution

Let u=1−x2.u = 1 - x^2. Then du=−2xdx.du = -2x\,dx.

Rewrite the integral: ∫xln⁡(1−x2)dx=−12∫ln⁡udu.\int x \ln(1-x^2)\,dx = -\frac12 \int \ln u \, du.

Now integrate: ∫ln⁡udu=uln⁡u−u.\int \ln u \, du = u\ln u - u.

Therefore, −12(uln⁡u−u)+C.-\frac12 (u\ln u - u) + C.

Substitute back: −12[(1−x2)ln⁡(1−x2)−(1−x2)]+C\boxed{ -\frac12\bigl[(1-x^2)\ln(1-x^2) - (1-x^2)\bigr] + C }

Original worksheet page 2: question and worked solution for 5-3-010

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