More Substitution Rule — Question 9

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Question 9

Evaluate the integral ∫x+1(x2+2x+3)3/2dx.\int \frac{x+1}{(x^2+2x+3)^{3/2}}\,dx.

Original worksheet page 1: question and worked solution for 5-4-009
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Question 9 - Solution

Begin by completing the square in the denominator: x2+2x+3=(x+1)2+2.x^2+2x+3=(x+1)^2+2.

Let u=x+1.u=x+1. Then du=dx,(x2+2x+3)3/2=(u2+2)3/2.du=dx, \qquad (x^2+2x+3)^{3/2}=(u^2+2)^{3/2}.

Substitute into the integral: ∫u(u2+2)3/2du.\int \frac{u}{(u^2+2)^{3/2}}\,du.

Now use a direct substitution. Let w=u2+2,dw=2udu.w=u^2+2, \qquad dw=2u\,du.

Then ∫u(u2+2)3/2du=12∫w−3/2dw.\int \frac{u}{(u^2+2)^{3/2}}\,du = \frac12\int w^{-3/2}\,dw.

Integrate: 12∫w−3/2dw=12(−2w−1/2)=−1w.\frac12\int w^{-3/2}\,dw = \frac12\left(-2w^{-1/2}\right) = -\frac{1}{\sqrt{w}}.

Substitute back: −1x2+2x+3+C\boxed{ -\frac{1}{\sqrt{x^2+2x+3}} + C }

Original worksheet page 2: question and worked solution for 5-4-009

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