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Partial fractions

by nameVoid
Tags: fractions, partial
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nameVoid
#1
Mar15-10, 04:24 AM
P: 247
[tex]\int \frac{2x+1}{4x^2+12x-7}dx[/tex]
[tex]\frac{1}{4} \int \frac{2x+1}{x^2+3x-\frac{7}{4}}dx[/tex]
[tex]\frac{1}{4} \int \frac{2x+1}{(x+\frac{3}{2})^2-4}dx[/tex]
[tex]u=x+\frac{3}{2}[/tex]
[tex]\frac{1}{2} \int \frac{u-1}{u^2-4}du[/tex]
[tex]\frac{1}{2} \int \frac{u}{u^2-4}du -\frac{1}{2}\int \frac{du}{u^2-4}[/tex]
[tex]\frac{1}{4} ln|u^2-4|-\frac{1}{2}\int \frac{A}{u+2} +\frac{B}{u-2} du[/tex]
[tex]-\frac{1}{2}=A(u-2)+B(u+2)[/tex]
[tex]A=\frac{1}{8}[/tex]
[tex]B=-\frac{1}{8}[/tex]
[tex]\frac{1}{4} ln|u^2-4|+\frac{1}{8}ln|\frac{u+2}{u-2}|+C[/tex]
[tex]\frac{1}{4} ln|x^2+3x-\frac{7}{4}|-\frac{1}{8} ln|\frac{x+\frac{7}{2}}{x-\frac{1}{2}}|+C[/tex]
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gabbagabbahey
#2
Mar15-10, 09:19 AM
HW Helper
gabbagabbahey's Avatar
P: 5,003
What you've got is correct, but it can still be simplified further; [itex]\ln|u^2-4|=\ln|u-2|+\ln|u-2|[/itex] and [itex]\ln\left|\frac{u+2}{u-2}\right|=\ln|u-2|-\ln|u-2|[/itex].

So,

[tex]\frac{1}{4}\ln|u^2-4|+\frac{1}{8}ln\left|\frac{u+2}{u-2}\right|=\frac{3}{8}\ln|u+2|+\frac{1}{8}\ln|u-2|[/tex]


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