Question

hi can someone help me with partial fractions integration lol

Answer 1

do you have a specific question or you want to know how the method works in general ?

Answer 2

no um i was working on a problem and got the wrong answer but don't see where i made the error can you check it for me ? lol

Answer 3

show us

Answer 4

\[\frac{4x^2+2x-1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\]

Answer 5

Oh...

Answer 6

wait are you sure lol

Answer 7

yeah

Answer 8

we have a repeated linear factor

Answer 9

Okay ill try that then lol thanks ill let you know!

Answer 10

Another possible way to do it is:
\[\frac{ Ax + B }{ x^2}+\frac{ C }{ x+1 }\]

Answer 11

Ill try both ways thanks @williamnz and @freckles

Answer 12

Okay so um im stuck here 4x^2+2x-1 = A(x^2)(x+1)+B(x)(x+1)+c(x)(x+1)---> x^2(A+B+C) +x(B+C) --> 4=A+B+C 2=B+C 0=Ax^3

Answer 13

\[4x^2+2x-1=Ax(x+1)+B(x+1)+Cx^2\]

Answer 14

Omg im so dumb i didn;t see that thanks!!!

Answer 15

\[\frac{4x^2+2x-1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\ \frac{A}{x} \frac{x(x+1)}{x(x+1)}+\frac{B}{x^2} \frac{x+1}{x+1}+\frac{C}{x+1} \frac{x^2}{x^2}\]

Answer 16

Hold on I think I got this

Answer 17

would it be Integral of 3/x + 1/x^2 + 1/(x+1) ??

Answer 18

is B=-1?

Answer 19

yes

Answer 20

oh oops

Answer 21

so you have 3/x-1/x^2+1/(x+1)

Answer 22

then we integrate that

Answer 23

yes i got that okay so um hold on

Answer 24

so is it 3ln|x| -1/x + ln|x+1| ???

Answer 25

first and last term great

Answer 26

i think the sign for the middle term if off

Answer 27

is* not if

Answer 28

Oh sorry its no negative lol

Answer 29

am i right now ?

Answer 30

looks good

Answer 31

+C

Answer 32

Thanks!! its not simplified tho right ?

Answer 33

to me it is

Answer 34

you could write it another way but to me its fine as is

Answer 35

Oh okay cause the answer is different to the book but thats because its simplified i think lol

Answer 36

yeah i think its just written a different way lol

Answer 37

\[3\ln|x|+\frac{1}{x}+\ln|x+1|+C \\ \ln|x^3|+\frac{1}{x}+\ln|x+1|+C \\ \ln|x^3(x+1)|+\frac{1}{x}+C \\ \ln|x^4+x^3|+\frac{1}{x}+C\]

Answer 38

the answer in the book says 1/x + ln|x^4+x^3|+c

Answer 39

isn't that what I wrote above

Answer 40

a+b =
b+a

Answer 41

lol i can't follow that but is it the same as what i have ?

Answer 42

addition is commutative

Answer 43

I just used properties of log

Answer 44

so i got the answer right just without adding them ? lol

Answer 45

power rule aln(x)=ln(x^a)
product rule ln(ab)=ln(a)+ln(b)