Question

evaluate the integral of (x^2+7x-28)dx/(x-2)^2(2x+1)

Answer 1

icky partial fraction problem. have to write is as
\[\frac{a}{x-2}+\frac{bx+c}{(x-2)^2}+\frac{d}{2x+1}\] and solve then you can integrate easily, but this part is a drag. it is cooked so you get nice integer solutions though

Answer 2

i am assuming the problem is
\[\int \frac{(x^2+7x-28)dx}{(x-2)^2(2x+1)}\]

Answer 3

yes

Answer 4

do you know how to get the partial fractions?

Answer 5

yes but i dont know how to solve for Bx+C

Answer 6

start with
\[a(x-2)(2x+1)+(bx+c)(2x+1)+d(x-2)^2=x^2+7x-28\]\]

Answer 7

you can get d right away by replacing x by -1/2

Answer 8

i have d but thats it

Answer 9

oh wait i told you wrong. it is not bx + c it is just b
in other words it looks like
\[\frac{a}{x-2}+\frac{b}{(x-2)^2}+\frac{c}{2x+1}\] because of the repeated root. then write
\[a(x-2)(2x+1)+b(2x+1)+c(x-2)^2=x^2+7x-28\]

Answer 10

you should get "d" which i am now calling "c" as -5 right?

Answer 11

i got C=5. i got x=1 and there aren't any subtraction signs when i use the heaviside method

Answer 12

i was told Ax+B for quadratic factors

Answer 13

yes but you only need a for a repeated root.

Answer 14

if you had
\[x^2+3x+1\] for example you would need
\[\frac{ax+b}{x^2+3x+1}\]

Answer 15

ok some how i messed up so let me start again. the idea is you multiply out and equate like coefficients

Answer 16

ok now i think i have it the easy way. start with
\[a(x-2)(2x+1)+b(2x+1)+c(x-2)^2=x^2+7x-28\] let x = 2 get
\[5b=-10\]
\[b=-2\] so now we have
\[a(x-2)(2x+1)-2(2x+1)+c(x-2)^2=x^2+7x-28\]
\[a(2x^2-3x-2)-4x-2+c(x^2-4x+4)=x^2+7x-28\]
\[2ax^2-3ax-2a-4x-2+cx^2-4cx+4c=x^2+7x-28\]
\[(2a+c)x^2+(-4c-2a-2)x-2a+4c-2=x^2+7x-28\]
so we know
\[2a+c=1\]
\[-2a+4c=-26\] add to get
\[5c=-25\]
\[c=-5\] then get "a"