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Small Algebra Problem

∫ ⁣a−x1+axdx=∫ ⁣−a−1+a2+1a[br](1+ax)dx\int \!{\frac {a-x}{1+ax}}{dx}=\int \!-{a}^{-1}+{\frac {{a}^{2}+1}{a[br] \left( 1+ax \right) }}{dx}

Hi all

I was wondering if you could explain the logical step that went into this algrbaic maniplulation. I understand could (infact) solve the integral a different way to this. But what has actually happened here?

Thanks

James Air
Reply 1
Avatar for DFranklin
DFranklin
18
a(a-x) = a^2 - ax = a^2+1-1-ax = (a^2+1)-(1+ax).

So (a-x) = [-(1+ax) + (a^2+1)] / a.

Now use that expression to replace (a-x) in the original integral.

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