help with inegral of 1/(1-x)

[Rohit93]

hi,
I was going through some solutions, i came across the integral
1/1-x
which the solutions shows as:
-log(1-x) + c
I know the integral of 1/x is log x +c so according to that the integral should be just log (1-x) + c right????. I cant get around the negative sign that appears. Please help

One more request as I am new to this forum can anyone please tell me how do i insert the integral and other symbols that would be very helpfull.
thanks

 

[HallsofIvy]

hi,
I was going through some solutions, i came across the integral
1/1-x
which the solutions shows as:
-log(1-x) + c
I know the integral of 1/x is log x +c so according to that the integral should be just log (1-x) + c right????. I cant get around the negative sign that appears. Please help

One more request as I am new to this forum can anyone please tell me how do i insert the integral and other symbols that would be very helpfull.
thanks

Strictly speaking, what you wrote was (1/1)- x= 1- x but I am sure you meant 1/(1- x). Please use parentheses!

Let u= 1- x so that du= -dx. Then \(\displaystyle \int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du\)
Can you do that integral?

To enter formulas use "Latex". for example, to do the integral above I wrote
"\int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du"
but started with [ tex ] and ended with [ /tex ] (without the spaces- I had to put those in so the "tex" would show up.)
There is a tutorial at
http://www.cs.cornell.edu/info/misc/latex-tutorial/latex-home.html

 

[Rohit93]

Strictly speaking, what you wrote was (1/1)- x= 1- x but I am sure you meant 1/(1- x). Please use parentheses!

Let u= 1- x so that du= -dx. Then \(\displaystyle \int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du\(\displaystyle
Can you do that integral?

To enter formulas use "Latex". for example, to do the integral above I wrote
"\int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du"
but started with [ tex ] and ended with [ /tex ] (without the spaces- I had to put those in so the "tex" would show up.)
There is a tutorial at
http://www.cs.cornell.edu/info/misc/latex-tutorial/latex-home.html\)\)

\(\displaystyle \(\displaystyle
Yeah my bad not using parentheses.
\(\displaystyle \int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du\)
If i do the above i get the following:
\(\displaystyle =-log(u)+c\)
and then resubstituting u=(1-x) i will get my desired answer correct????
Thanks for the "LATEX" link it has been a great help\)\)

 

[lookagain]

strictly speaking, what you wrote was (1/1)- x= 1- x but i am sure you meant 1/(1- x). Please use parentheses!

Let u= 1- x so that du= -dx. Then \(\displaystyle \int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du \ \ \ \ \ \)<----- the "/" was left out next to "tex" so the latex didn't show.
Can you do that integral?

To enter formulas use "latex". For example, to do the integral above i wrote
"\int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du"
but started with [ tex ] and ended with [ /tex ] (without the spaces- i had to put those in so the "tex" would show up.)
there is a tutorial at
http://www.cs.cornell.edu/info/misc/latex-tutorial/latex-home.html

\(\displaystyle \int \frac{1}{1- x}= \int \frac{1}{u}(-du)= -\int \frac{1}{u}du\)
If i do the above i get the following:
\(\displaystyle =-log(u)+c\)

Rohit93,

it equals -log|u| + C, where "log" is the natural logarithm, or -ln|u| + C, before substituting back for x.

You have to have the absolute value bars

 

[stapel]

I came across the integral 1/(1-x) which the solutions shows as:

-log(1-x) + c

I know the integral of 1/x is log x +c so according to that the integral should be just log (1-x) + c right? I can't get around the negative sign that appears.

The integral of 1/x is ln(x) + c. The integral of -1/x is, naturally, -ln(x) + c. You are starting with an integrand stated as 1/(1 - x). This may be restated, using the fact that 1 - x = -x + 1 = -(x - 1), as -[1/(x - 1)]. By "flipping" the subtraction, the integrand better matches the original pattern you noted. Now let u = x - 1, so du = dx, and see what happens.

Otherwise, work clearly and completely with the u-substitution. If u = 1 - x, then du = -dx, and so forth. By showing all of the work and reasoning, the logic of the answer becomes clear.

 

[Rohit93]

The integral of 1/x is ln(x) + c. The integral of -1/x is, naturally, -ln(x) + c. You are starting with an integrand stated as 1/(1 - x). This may be restated, using the fact that 1 - x = -x + 1 = -(x - 1), as -[1/(x - 1)]. By "flipping" the subtraction, the integrand better matches the original pattern you noted. Now let u = x - 1, so du = dx, and see what happens.

Otherwise, work clearly and completely with the u-substitution. If u = 1 - x, then du = -dx, and so forth. By showing all of the work and reasoning, the logic of the answer becomes clear.

Stapel,
Thank you it is now it is clearer this way!!!!!!

 

[lookagain]

The integral of 1/x is ln(x) + c. The integral of -1/x is, naturally, -ln(x) + c.

\(\displaystyle \displaystyle\int\dfrac{1}{x}dx \ = \ ln|x| \ + \ C\)


\(\displaystyle \displaystyle\int \bigg(-\dfrac{1}{x}\bigg)dx \ = \ -ln|x| \ + \ C\)