+0

intergration of x to the minus one.

+3
704
5

I know this is lnx,but how is this arrive at?

Jul 2, 2014

#2
+11

Here's another way of looking at it (but also by differentiating lnx): Jul 2, 2014

#1
+8

Let us start by letting  f(x) = e^x and g(x) = lnx

And these functions are inverses of each other.

There is  a fact associated with inverse functions that says that:

g'(x) = 1 / f '(g(x))     ..... so using this, we have

g'(x) = 1/ e^(g(x)) = 1/ e^(lnx) = 1/x   (remember that f'(x) = e^x)

Putting this together, we have

(d/dx)[ ln(x) + C] = 1/x      (the derivative of the constant "C" = 0)

Thus, taking the derivative of the anti-derivative ln(x) + C  returns the function in the integrand, i.e., 1/x = x^(-1)   Jul 2, 2014
#2
+11

Here's another way of looking at it (but also by differentiating lnx): Alan Jul 2, 2014
#3
0

to add on to this i want to mention that a lot of people try to use the antiderivative equivalent of the power rule where we add 1 to the exponent and then divide by it.  I saw this a lot with my own students this year.  they would do this:

(x^-1+1)/(-1+1)=(x^0)/0   and then they would try to tell me the answer was 0.  what in the world!  This is obviously undefined

Jul 2, 2014
#4
+5

It should also be noted that the antiderivative is ln|x|+c and yes it is very important that we include the absolute value bars in our answer.  The function f(x)=1/x has all real numbers, with the exception of 0, as its domain so naturally we want the antiderivative to have the same domain.  Notice that ln(x) has a domain that includes on the positive real numbers.  In when we use the absolute value inside the natural log it will now accept negative real numbers and inputs so now it has the same domain as 1/x.

Jul 2, 2014
#5
+6

$$\boxed{{\ d{(\ln{y})} \over \ dx }={y' \over y }}\\\\ \Rightarrow {\ d{(\ln{x})} \over \ dx }={1 \over x }\\\\ \Rightarrow {\ d({\ln{(a\times x)}}) \over \ dx } ={a \over a\times x } ={1 \over x }$$

$$\begin{array}{ccl} \int{ {1\over x} \ dx }&=& \ln{(|a\times x|)} +c_1\\ & = &\ln{(|a|)}+\ln{(|x|)}+c_1 \\ & = &\ln{(|x|)}+\underbrace{\ln{(|a|)}+c_1}_{c_2} \\ & = &\ln{(|x|)}+c_2} \end{array}$$

.
Jul 3, 2014