Integrals Involving Roots — Question 2

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Question 2

Use u=x+2u=x+2 to evaluate: ∫xx+2dx.\int x\sqrt{x+2}\,dx.

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

Step 1: Substitute u=x+2u=x+2. Then x=u−2,du=dx.x=u-2,\qquad du=dx. Step 2: Rewrite and expand the integrand. ∫xx+2dx=∫(u−2)u1/2du=∫(u3/2−2u1/2)du.\begin{align*} \int x\sqrt{x+2}\,dx &=\int (u-2)u^{1/2}\,du\\ &=\int\left(u^{3/2}-2u^{1/2}\right)du. \end{align*} Step 3: Integrate each power. I=u5/25/2−2u3/23/2+C=25u5/2−43u3/2+C.\begin{align*} I&=\frac{u^{5/2}}{5/2}-2\frac{u^{3/2}}{3/2}+C\\ &=\frac25u^{5/2}-\frac43u^{3/2}+C. \end{align*} Step 4: Substitute u=x+2u=x+2. 25(x+2)5/2−43(x+2)3/2+C\boxed{\frac25(x+2)^{5/2}-\frac43(x+2)^{3/2}+C}

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

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