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## Main Question or Discussion Point

integral dx/sqrt(x - a)

integral dx/sqrt(1/ax)

integral dx/sqrt(1/ax)

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integral dx/sqrt(x - a)

integral dx/sqrt(1/ax)

integral dx/sqrt(1/ax)

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reading suggestions

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- #4

HallsofIvy

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We **all** embarrass ourselves at one time or another! (Some of us more than others. Believe me, I **know** about embarrassing myself!) **PLEASE** come back to PhysicsForums!

(I need someone to make**me** look good!!!)

Anyway, for those who are still wondering how to do these:

[tex]\int \sqrt{x-a}dx[/tex]

Let u= x-1 so du= dx and [tex]\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}[/tex]. The integral becomes [tex]\int u^{\frac{1}{2}}du[/tex] which can be done with the power law.

[tex]\int \frac{dx}{\sqrt{\frac{1}{ax}}}[/tex]

[tex]\frac{1}{\sqrt{\frac{1}{ax}}}[/tex] is just [tex]\sqrt{ax}[/tex].

Let u= ax so du= adx, dx= (1/a)du and [tex]\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}[/tex]

The integral becomes [tex]\frac{1}{a}\int u^{\frac{1}{2}}du[/tex]

(I need someone to make

Anyway, for those who are still wondering how to do these:

[tex]\int \sqrt{x-a}dx[/tex]

Let u= x-1 so du= dx and [tex]\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}[/tex]. The integral becomes [tex]\int u^{\frac{1}{2}}du[/tex] which can be done with the power law.

[tex]\int \frac{dx}{\sqrt{\frac{1}{ax}}}[/tex]

[tex]\frac{1}{\sqrt{\frac{1}{ax}}}[/tex] is just [tex]\sqrt{ax}[/tex].

Let u= ax so du= adx, dx= (1/a)du and [tex]\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}[/tex]

The integral becomes [tex]\frac{1}{a}\int u^{\frac{1}{2}}du[/tex]

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HallsofIvy said:[tex]\int \sqrt{x-a}dx[/tex]

HallsofIvy, should not this equation be:

[tex]\int \frac{dx}{\sqrt{x - a}} = \int u^{-\frac{1}{2}} du[/tex]

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HallsofIvy

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