Question

(x^2+3x)/(x^2-4) partial fraction with steps please

Answer 1

\[\frac{x^2+3x}{x^2-4}\equiv\frac{A}{x-2}+\frac{B}{x+2}\]

Answer 2

I keep getting 1-5/2(x+2)-1/2(x-2)

Answer 3

wait, i mean:
\[\frac{x^2+3x}{x^2-4}\equiv x\left(\frac A{x-2}+\frac B{x+2}\right)\]

Answer 4

I don't know what I'm doing wrong

Answer 5

Ax^2+Bx^2=x^2
2Ax-2Bx=3x

Answer 6

A+B=1
2A-2B=3

Answer 7

2(1)-(2): 4B=-1, B=-0.25

Answer 8

(1): A=1.25

Answer 9

\[\frac{x^2+3x}{x^2-4}\equiv\frac{1.25x}{x-2}-\frac{0.25x}{x+2}\]

Answer 10

what did you get? please use proper brackets

Answer 11

The answer says 1+(5/2)(x-2)+1/2(x+2) but I get 1-(5/2)(x+2)-1/2(x-2)

Answer 12

wait, can you use a divide sign to separate the numerator and the denominator

Answer 13

i can hardly interpret what you wrote

Answer 14

5/2 is a fraction as well as 1/2

Answer 15

you need to divide first to make denominator's degree less than numerator

Answer 16

I did that

Answer 17

Hence the 1

Answer 18

Do you know what you wrote
\[1+\frac52(x-2)+\frac12(x+2)\]

Answer 19

\(\large \frac{x^2+3x}{x^2-4} = \frac{x^2-4 + 4 + 3x}{x^2-4} = 1 + \frac{3x+4}{x^2-4}\)

Answer 20

Yes I got to that part

Answer 21

Wait almost

Answer 22

\(\large \frac{3x+4}{x^2-4} = \frac{A}{x-2} + \frac{B}{x+2}\)

use cover up and find A, B

Answer 23

I think the signs are different

Answer 24

That's my mistake

Answer 25

The answer says 1+(5/2)(x-2)+1/2(x+2) but I get 1-(5/2)(x+2)-1/2(x-2)
^^^^^ ^^^^^

Answer 26

work it again, not just signs.. values above A and B also flipped

Answer 27

Okay I believe I've done the division in the first step in reverse
Thank you

Answer 28

cool ^_^

Answer 29

Now I can integrate

Answer 30

yes, that should be trivial, just the ln 's

Answer 31

:)