Integral of (1-x)/x^2: -(1/x) + ln(x) + c?

[m1ndpixel]

Hi,

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

 

[rock.freak667]

Hi,

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

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

 

[m1ndpixel]

sorry, yes that's what i meant!

 

[rock.freak667]

sorry, yes that's what i meant!

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

 

[HallsofIvy]

Hi,

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

Almost. (1-x)/x²= 1/x²- 1/x and the integral of that is -1/x- ln(x)+ c

Mathematica is giving some strange answer containing a log.
Thanks

??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.