More Substitution Rule — Question 1

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Question 1

Evaluate the integral ∫x1+x21+ln⁡(1+x2)dx.\int\frac{x}{1+x^2}\sqrt{1+\ln(1+x^2)}\,dx.

Original worksheet page 1: question and worked solution for 5-4-001
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Question 1 - Solution

Step 1: Choose a substitution

Set u=1+ln⁡(1+x2)u=1+\ln(1+x^2), so du=2x/(1+x2)dxdu=2x/(1+x^2)\,dx.

∫x1+x21+ln⁡(1+x2)dx=12∫u1/2du=13u3/2+C.\int\frac{x}{1+x^2}\sqrt{1+\ln(1+x^2)}\,dx =\frac12\int u^{1/2}\,du=\frac13u^{3/2}+C.

13(1+ln⁡(1+x2))3/2+C.\boxed{\frac13\bigl(1+\ln(1+x^2)\bigr)^{3/2}+C.}

This is valid for every real xx.

Original worksheet page 2: question and worked solution for 5-4-001

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