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OpenStudy (anonymous):
integrate 4x-7/x^2-x-6 dx
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OpenStudy (anonymous):
\[\int\limits_{}^{}4x-7/x ^{2}-x-6 dx\]
OpenStudy (amistre64):
\frac{n}{d} if your doing the latex coding
OpenStudy (amistre64):
(x-3)(x+2) on the denom, then partial it
OpenStudy (amistre64):
it should result in: A ln(x-3) + Bln(x+2)
where A and B are the numerators from the partials
OpenStudy (anonymous):
then u set it = to the numerator
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OpenStudy (amistre64):
yep
OpenStudy (anonymous):
ok, i'll try that. thank you!
OpenStudy (amistre64):
the "cover up" technique might be simplest
OpenStudy (amistre64):
\[\frac{4x-7}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}\]
\[when\ x=3:\ \frac{4.3-7}{\cancel{(3-3)}(3+2)}=\frac{A=8/5}{x-3}+\frac{B}{x+2}\]
\[when\ x=-2:\ \frac{4.-2-7}{(-2-3)\cancel{(-2+2)}}=\frac{8/5}{x-3}+\frac{B=-15/-5}{x+2}\]
OpenStudy (amistre64):
12-7=5 ; not 8 :)
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OpenStudy (anonymous):
ok. so far i got B=3
OpenStudy (amistre64):
yep
OpenStudy (anonymous):
and A=1
OpenStudy (amistre64):
A=5/5 = 1 is what i get when i learn to multiply
OpenStudy (anonymous):
ok. i got A=1 too
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OpenStudy (anonymous):
i got ln(x-3)+3ln(x+2)
OpenStudy (anonymous):
i had a quick question. how do you know when to use partial fraction decomposition?
OpenStudy (amistre64):
if the top is not the derivative of the bottom, we look to arctan or partials
OpenStudy (anonymous):
oh ok. Thank you :)
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