Question

Integral of (3x-2)/(x-1)^2

Answer 1

Using partial fractions Idk how to successfully get my A value

Answer 2

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html

Answer 3

\[\frac{3x-2}{(x-1)^2}=\frac{A}{x-1}+\frac{Bx}{(x-1)^2}=\frac{A(x-1)+Bx}{(x-1)^2}\]
3x-2=A(x-1)+Bx
3x-2=x(A+B)-A
A+B=3
-A=-2=>A=2
A+B=3=>2+B=3=>B=1
lets check our partial fraction
we have
\[\frac{2}{x-1}+\frac{x}{(x-1)^2}=\frac{2(x-1)+x}{(x-1)^2}=\frac{3x-2}{(x-1)^2}\]
so yay no mistakes!

Answer 4

Why is it Bx and not just A + B

Answer 5

it can be just B since they are mulitples of linears

Answer 6

but it works out anyways right?

Answer 7

Do you mean it can or can't?

Answer 8

when in doubt; try it both ways ... unless its illegal :)

Answer 9

And it is just B.

Answer 10

\[\int\limits_{}^{}\frac{3x-2}{(x-1)^2}dx=\int\limits_{}^{}(\frac{2}{x-1}+\frac{x}{(x-1)^2})dx=2\ln|x-1|+\int\limits_{}^{}\frac{x}{(x-1)^2}dx\]
let u=x-1
so du=1dx
also if u=x-1, then x=u+1
so we have
\[2\ln|x-1|+\int\limits_{}^{}\frac{u+1}{u^2}du=2\ln|x-1|+\int\limits_{}^{}(\frac{1}{u}+\frac{1}{u^2})du=2\ln|x-1|+\ln|u|-\frac{1}{u}\+C]
=\[2\ln|x-1|+\ln|x-1|-\frac{1}{x-1}+C=\ln|x-1|^2+\ln|x-1|-\frac{1}{x-1}+C\]
=\[\ln|{(x-1)^2(x-1)}|-\frac{1}{x-1}+C=\ln|(x-1)^3|-\frac{1}{x-1}+C\]
you can use this to check yourself after you do the problem if you like
Check: \[(\ln|(x-1)^3|-\frac{1}{x-1}+C)'=3*\frac{1}{x-1}+\frac{1}{(x-1)^2}+0=\frac{3(x-1)+1}{(x-1)^2}=\frac{3x-1+1}{(x-1)^2}
=\frac{3x}{(x-1)^2}\]
so i made a mistake somewhere

Answer 11

my thingy got cut off