Integral with partial fractions....
\[\int\limits 3/x^3-9x^2 dx\]
.....\[a= -3/x B= -1/3/x^2 C= 3/x-9 \]
when I take integral I get.... \[-3\ln(x)-1/3\ln(x^2)+3\ln(x-9)\]
Does this look correct?
\[\frac{ 3 }{ x^{2}(x-9) }= \frac{ A }{ x }+ \frac{ B }{ x^{2} }+\frac{ C }{ x-9 }\] \[3=Ax(x-9)+B(x-9)+Cx^{2}\] If x = 0, B = -1/3 If x = 9, C = 1/27 If x = 10 , A = -1/27 \[\int\limits_{}^{}-\frac{ 1 }{ 27x }-\frac{ 1 }{ 3x^{2} }+\frac{ 1 }{ 27(x-9) } = -\frac{ 1 }{ 27 }\ln|x|+\frac{ 1 }{ 3x }+\frac{ 1 }{ 27 } \ln|x-9| +C\] this is what I get
So when getting to A is part of it -1/3/9?
which simplifies to -1/27?
Given we already solved for B and C, I choose x = 10 to make everything as simplified as possible: \[3 = 10A- \frac{ 1 }{ 3 }+ \frac{ 100 }{ 27 }\]I make the 3 on the left and the -1/3 on the right into common denominators of 27 \[\frac{ 81 }{ 27 }= 10A- \frac{9}{27} + \frac{100}{27} \implies \frac{81}{27} = 10A + \frac{91}{27} \implies \frac{-10}{27} = 10A \] Then dividing by 10 gets the A I got before.
ok.... I was at a point though where I had (-1/3)/9 Can that simplify to -1/27th? If so I just got that part mixed up.
That can become -1/27, yeah. Just was showing you how I got there.
and get to the same answer ln((x-9)/x)^1/27 + 1/3x + C
Thank you so much for your help! A few of the general steps are more clear now (;
Yeah, np : )
Are you in school?
Yeah, Im still a college student.
What school and major?
Part time at a community college and UNLV. Math major, already have an associates in Japanese, too, and now its my minor at UNLV.
Awesome! Thanks again for the help. Im at LCC Oregon, working on an engineering transfer degree to OSU.
Cool, lol. Im still stuck here for a little bit, but Im planning to transfer to a university in Tokyo. Temple has their own overseas branch in Tokyo, so thats the plan come fall of 14.
I bet that will be fun for you!
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