- #1
John O' Meara
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Integrate [tex] \int \frac{x}{sqrt{x+3}}dx \\ [/tex]
(i) I can integrate this by letting U^2=x+3 => [tex] u=\sqrt{x+3} \ \mbox{ so } \ x=u^2-3 \ \mbox{and } \ dx=2udu \ \mbox{then the integral = } \ \int\frac{(u^2-3)2u}{u}du \\[/tex]
[tex] \ =2(\frac{u^3}{3}-3u) + C \\ \ = \ \frac{2}{3}(u^2-9) + C \ = \ \frac{2}{3}(x-6)\sqrt{x+3} +C \\ [/tex]
Or (ii) I can integrate above if I let u=x+3 => [tex] \ \frac{du}{dx}=1 \ \mbox{ therefore } \ du = dx \ \mbox{ and } \ x=u-3 \ \\[/tex]. [tex] \mbox{ then } = \ \int\frac{u-3}{u^{\frac{1}{2}} du \ = \ \int u^{\frac{1}{2}}du \ -\ 3 \int u^{\frac{-1}{2}} du \ = \ \frac{2}{3} u^{\frac{3}{2}} \ - \ 6u^{\frac{1}{2}} +C \ \\[/tex].[tex] \ = \ \frac{2}{3}(x+3)^{\frac{3}{2}} \ -6(x+3)^{\frac{1}{2}} + C\\[/tex]. Which is correct (i) or (ii), and why? Thanks alot.
(i) I can integrate this by letting U^2=x+3 => [tex] u=\sqrt{x+3} \ \mbox{ so } \ x=u^2-3 \ \mbox{and } \ dx=2udu \ \mbox{then the integral = } \ \int\frac{(u^2-3)2u}{u}du \\[/tex]
[tex] \ =2(\frac{u^3}{3}-3u) + C \\ \ = \ \frac{2}{3}(u^2-9) + C \ = \ \frac{2}{3}(x-6)\sqrt{x+3} +C \\ [/tex]
Or (ii) I can integrate above if I let u=x+3 => [tex] \ \frac{du}{dx}=1 \ \mbox{ therefore } \ du = dx \ \mbox{ and } \ x=u-3 \ \\[/tex]. [tex] \mbox{ then } = \ \int\frac{u-3}{u^{\frac{1}{2}} du \ = \ \int u^{\frac{1}{2}}du \ -\ 3 \int u^{\frac{-1}{2}} du \ = \ \frac{2}{3} u^{\frac{3}{2}} \ - \ 6u^{\frac{1}{2}} +C \ \\[/tex].[tex] \ = \ \frac{2}{3}(x+3)^{\frac{3}{2}} \ -6(x+3)^{\frac{1}{2}} + C\\[/tex]. Which is correct (i) or (ii), and why? Thanks alot.
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