I'm making a small mistake somewhere, but I can't seem to find it.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int\frac{dx}{(x-1)(1-2x)}[/tex]

taking the partial fractions

[tex]1=A(1-2x)+B(x-1)[/tex]

[tex]A=-1, B=-2[/tex]

[tex]\int\frac{-1}{x-1} dx+\int\frac{-2}{1-2x}dx[/tex]

Integrating by substitution, this is what I'm getting

[tex]-ln(x-1)+ln(1-2x)+C[/tex]

The correct answer is

[tex]-ln(x-1)+ln(2x-1)+C[/tex]

I think I'm making some kind of algebraic sign mistake when taking the second integral, but I just can't find it. I'll go through my solution step by step here and if someone could point out the mistake I would appreciate it.

[tex]\int\frac{-2}{1-2x}dx[/tex]

[tex]-2\int\frac{dx}{(1-2x)}[/tex]

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

[tex]-2\int\frac{dx}{-2u}[/tex]

[tex]\int\frac{dx}{u}[/tex]

[tex]ln(u)+C[/tex]

[tex]ln(1-2x)+C[/tex]

Somewhere along the way, I'm missing the spot where I'm supposed to factor out a -1 from the denominator, but where?

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# Homework Help: Integration by Partial Fractions

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