- #1

- 2

- 0

The integral of ((1-x)/x^2)dx is: -(1/x) + ln(x) + c

Is that correct?

Mathematica is giving some strange answer containing a log.

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter m1ndpixel
- Start date

- #1

- 2

- 0

The integral of ((1-x)/x^2)dx is: -(1/x) + ln(x) + c

Is that correct?

Mathematica is giving some strange answer containing a log.

Thanks

- #2

rock.freak667

Homework Helper

- 6,223

- 31

+ln(x) should be -ln(x)

The integral of ((1-x)/x^2)dx is: -(1/x) + ln(x) + c

Is that correct?

Mathematica is giving some strange answer containing a log.

Thanks

- #3

- 2

- 0

sorry, yes thats what i meant!

- #4

rock.freak667

Homework Helper

- 6,223

- 31

sorry, yes thats what i meant!

then it should be correct providing that you remember to change the + to a -.

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

Almost. (1-x)/xHi,

The integral of ((1-x)/x^2)dx is: -(1/x) + ln(x) + c

Is that correct?

??YOUR answer contains a log! What was the strange answer Mathematica gave? If you mean that Mathematica is giving an answer with a logarithm base 10, perhaps it is using the fact that ln(x)= log(x)/log(e) where "log" is logarithm base 10. If it is just giving -1/x- log(x)+ c, then it is using "log" to mean natural logarithm. That is fairly standard now where common logarithms are not much used.Mathematica is giving some strange answer containing a log.

Thanks

Share: