Question

find the partial fraction decomposition

Answer 1

\[\int\limits_{?}^{?}\frac{ 2x }{ (x-1)^3 }\]

Answer 2

no i dont know how to start this one

Answer 3

is it going to look something like this

Answer 4

\[\int\limits \frac{ 2x }{ (x-1)^3 }dx=\int\limits \frac{ 2x-2+2 }{ (x-1)^3 }dx=2 \int\limits (x-1)^{-2}dx+2\int\limits (x-1)^{-3}dx+c\]

Answer 5

how did you get 2x-2+2?

Answer 6

-2+2=0 to make 2(x-1) in the numerator.

Answer 7

@surjithayer we dont use integrals on this one just go to find the answer using partial fraction decomposition

Answer 8

you are doing good,that is correct ,just put + sign in between.

Answer 9

@surjithayer is this right

Answer 10

\(2x=A(x-1)^2+B(x-1)+C\)

Answer 11

what about the (x-1)^3?

Answer 12

we multiply by (x-1)^3

Answer 13

i dont get it

Answer 14

comparing coefficients of like powers

Answer 15

\[\frac{ 2x }{ \left( x-1 \right)^3 }=\frac{ A }{ x-1 }+\frac{ B }{ \left( x-1 \right)^2 }+\frac{ C }{ \left( x-1 \right)^3 }\]

Answer 16

yes i got that

Answer 17

now multiply each term by \((x-1)^3\)

Answer 18

why are we doing that?

Answer 19

we always multiply each term by the denominator on left hand side ,then there is no term in the denominator,

Answer 20

the answer on the back of my book is 2/(x-1)^2+1/(x-1)^3 but i got C=2

Answer 21

i used 1 for x

Answer 22

you are correct.

Answer 23

but there is no c in the answer

Answer 24

my bad there is a c but its 1/(x-1)^3

Answer 25

while integrating we always add general constant say c.

Answer 26

@surjithayer i dont think i am doing it right

Answer 27

\(2x=A(x-1)^2+B(x-1)+C\)
equating coefficient of \(x^2\)
0=A
equating coefficient of x
2=-2A+B
B=2
equating constant term
0=-A-B+C
0=0-B+C
C=B
so C=2

Answer 28

correction for constant term
0=A-B+C

Answer 29

how does A=0

Answer 30

okay

Answer 31

\(2x=0 x^2+2x+0\)

Answer 32

alright