Integration Strategy — Question 3

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Question 3

Use u=x2+1u=x^2+1 to evaluate: ∫x3x2+1dx.\int\frac{x^3}{\sqrt{x^2+1}}\,dx.

Original worksheet page 1: question and worked solution for 1-7-003
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Question 3 – Solution

Step 1: Separate one factor of xx. x3dx=x2(xdx).x^3\,dx=x^2(x\,dx). Step 2: Substitute u=x2+1u=x^2+1. Then du=2xdx,xdx=12du,x2=u−1.du=2x\,dx,\qquad x\,dx=\frac12du,\qquad x^2=u-1. Step 3: Rewrite and simplify. I=12∫u−1udu=12∫(u1/2−u−1/2)du.\begin{align*} I&=\frac12\int\frac{u-1}{\sqrt{u}}\,du\\ &=\frac12\int\left(u^{1/2}-u^{-1/2}\right)du. \end{align*} Step 4: Integrate each power. I=12(u3/23/2−u1/21/2)+C=13u3/2−u1/2+C.\begin{align*} I&=\frac12\left(\frac{u^{3/2}}{3/2}-\frac{u^{1/2}}{1/2}\right)+C\\ &=\frac13u^{3/2}-u^{1/2}+C. \end{align*} Step 5: Return to xx. 13(x2+1)3/2−x2+1+C\boxed{\frac13(x^2+1)^{3/2}-\sqrt{x^2+1}+C}

Original worksheet page 2: question and worked solution for 1-7-003

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