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 ...

- anonymous

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- anonymous

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

- anonymous

\[1- \frac{1}{x+1}\]
that should be last part

- anonymous

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

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## More answers

- anonymous

can you give me another example?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

its all fairly standard

- anonymous

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?

- anonymous

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

- anonymous

in the numerator

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you cant break that up

- anonymous

partial fractions? idk you need a constant on top

- anonymous

but that cant be broken down

- anonymous

ya

- anonymous

you cant just make up random examples

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I dont understand whart u mean

- anonymous

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

- anonymous

then its a logarithm

- anonymous

yeah that is right

- anonymous

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

- anonymous

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

- anonymous

my question is how do I do that

- anonymous

ohh yeh, now you have changed the question :|

- anonymous

sorry lol

- anonymous

add and subtract 1 on the top and bottom

- anonymous

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

- anonymous

very basic, nothing really hard

- anonymous

nice

- anonymous

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

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