Question

if I have an expression like this x/(x+1). I can divide numerator by denominator and get something like this 1-1/(x+1). Can someone teach me how to ...

Answer 1

express the numerator as \[\frac{ x+1-1}{x+1} = \frac{ x+1}{x+1} + \frac{-1}{x+1} = 1 \frac{1}{x+1} \]

Answer 2

\[1- \frac{1}{x+1}\]

that should be last part

Answer 3

we add and subtract the same thing, so we dont change the value overall

Answer 4

can you give me another example?

Answer 5

take \[\frac{x+2}{x+3} \]

Answer 6

so we look on the bottom , it has x+3 , so we want to see an x+3 in the numerator

Answer 7

so

express as \[\frac{x+3+2 -3 }{x+3} = \frac{x+3 -1 }{x+3}\]

Answer 8

\[= \frac{x+3}{x+3} - \frac{1}{x+3} = 1 - \frac{1}{x+3}\]

Answer 9

its all fairly standard

Answer 10

so you mean I have to look at the bottom. if I have for example (x+4), then I have to add and substract 4 and -4 respectively?

Answer 11

yes , all it relys on is being able to split up the numerators

Answer 12

in the numerator

Answer 13

ok im gonna give you an example. Please check if it is correct

Answer 14

\[x/(1+x^{2})\]well i dont know what to do here :(

Answer 15

thats why it only works with linear terms in the numerator and denominator

Answer 16

you cant break that up

Answer 17

partial fractions? idk you need a constant on top

Answer 18

but that cant be broken down

Answer 19

ya

Answer 20

you cant just make up random examples

Answer 21

or you could have something like \[\frac{x^2 +2x} {(x+1)^2} \]

Answer 22

So how should I do in cases like this. that is why I dont understand The oiginal problems starts there

Answer 23

the question is badly written, I would just ignore it, it cant be broken down further , teacher is stupid

Answer 24

Sorry dude but im working with integrals. They have an expression like that and get that answer. the answer is 1-1(x2 +1)

Answer 25

I dont understand whart u mean

Answer 26

but if you want \[\int\limits \frac{x}{1+x^2} dx \]

Answer 27

then its a logarithm

Answer 28

yeah that is right

Answer 29

in a book im reading, they say that \[x ^{2}/(1+x ^{2})=1-1/(1+x ^{2})\]

Answer 30

\[= \frac{1}{2} \ln ( 1+x^2) +\]

Answer 31

my question is how do I do that

Answer 32

ohh yeh, now you have changed the question :|

Answer 33

sorry lol

Answer 34

add and subtract 1 on the top and bottom

Answer 35

\[=\frac{x^2+1-1}{x^2+1} = \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} = 1- \frac{1}{x^2+1}\]

Answer 36

very basic, nothing really hard

Answer 37

nice

Answer 38

ok cool. sorry for me to make you feel angry lol