Substitution Rule for Indefinite Integrals — Question 2

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Question 2

Evaluate the integral ∫ln⁡(x2+1)x2+1xdx.\int \frac{\ln(x^2+1)}{x^2+1}\,x\,dx.

Original worksheet page 1: question and worked solution for 5-3-002
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Question 2 - Solution

Use substitution. Let u=ln⁡(x2+1).u=\ln(x^2+1). Then du=2xx2+1dx⇒xx2+1dx=12du.du=\frac{2x}{x^2+1}\,dx \quad\Rightarrow\quad \frac{x}{x^2+1}\,dx=\frac{1}{2}\,du.

Substitute: ∫ln⁡(x2+1)x2+1xdx=12∫udu.\int \frac{\ln(x^2+1)}{x^2+1}\,x\,dx = \frac{1}{2}\int u\,du.

Integrate: 12∫udu=14u2.\frac{1}{2}\int u\,du=\frac{1}{4}u^2.

Substitute back: 14(ln⁡(x2+1))2+C\boxed{ \frac{1}{4}\bigl(\ln(x^2+1)\bigr)^2 + C }

Original worksheet page 2: question and worked solution for 5-3-002

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