Question

Can someone explain why \[\int\limits x ^{-1} dx\] is anti-derivative \[\int\limits \ln(-x) + C\]

Answer 1

it isn't

Answer 2

meaning?

Answer 3

\[\int \frac{dx}{x}=\ln(x)\]

Answer 4

well \[x^{-1} = 1/x\]?

Answer 5

\[\ln(-x)=\ln(\frac{1}{x})\] although i don't know if that helps any
the anti - derivative of \(x^{-1}\) is \(\ln(x)\)

Answer 6

i am confused by your question, because you have two indefinite integrals in it

Answer 7

how about this can you work out first equation

Answer 8

maybe

Answer 9

plz <3

Answer 10

ok what is it?

Answer 11

i don't get why if you use \[\frac{ x ^{n+1}}{ n+1 }\] that it would not just be \[x^0 + C\]

Answer 12

ok i think i understand the question
it is this:
what is\[\int x^{-1}dx\] or equivalently what is
\[\int \frac{1}{x}dx\] right?

Answer 13

since the exponent is \(-1\) why can't you just use the power rule backwards, i.e. us
\[\int x^ndx=\frac{x^{n+1}}{n+1}\]

Answer 14

it doesn't work in the case \(x^{-1}\) for a couple reasons, but one is this: if you used it literally you would get
\[\int x^{-1}dx=\frac{x^{-1+1}}{-1+1}=\frac{x^0}{0}\]

Answer 15

and of course you cannot divide by zero!

Answer 16

so that's is why i cant use the power rule backwards right?

Answer 17

yes, it is not applicable in the this case because the denominator is zero

Answer 18

now as to why
\[\int x^{-1}dx=\ln(x)\], that depends largely on your definition of the log
but one simple reason is that
\[\frac{d}{dx}[\ln(x)]=\frac{1}{x}\] so pretty much by definition,
\[\int \frac{1}{x}dx=\ln(x)\]

Answer 19

later on in this or some other math class you will DEFINE the log to be
\[\ln(x)=\int_1^x\frac{dt}{t}\]

Answer 20

at the moment my guess is your definition of the log is the "inverse of the exponential"

Answer 21

i sorta get it.

Answer 22

the derivative of \(\ln(x)\) is \(\frac{1}{x}\) so the anti derivative of \(\frac{1}{x}\) is \(\ln(x)\)
that should work for your calc class

Answer 23

i am assuming you know that the derivative of \(\ln(x)\) is \(\frac{1}{x}\) from before, rigth?

Answer 24

yes

Answer 25

i was just wondering why it would not be x^0 but you explain that very well

Answer 26

ok then that explanation should be sufficient, because if the derivative of \(F\) is \(f\) then the anti derivative of \(f\) is \(F\)

Answer 27

ok good

Answer 28

alright man thanks for the help

Answer 29

yw