+ln(x) should be -ln(x)m1ndpixel said: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
m1ndpixel said:sorry, yes that's what i meant!
Almost. (1-x)/x2= 1/x2- 1/x and the integral of that is -1/x- ln(x)+ cm1ndpixel said:Hi,
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
The integral of (1-x)/x^2 is -(1/x) + ln(x) + c.
To solve the integral of (1-x)/x^2, you can use the formula: integral of (1/x^n) = x^(-n+1)/(n-1) + c. In this case, n = 2, so the integral becomes -(1/x) + c + ln(x).
The constant c represents the value of the indefinite integral, which is the sum of all possible antiderivatives of the function. It is used to account for any possible constant terms in the original function.
The limits of integration for the integral of (1-x)/x^2 depend on the specific problem or context in which the integral is being used. Generally, the limits are chosen to correspond to the boundaries of the function being integrated.
Yes, the integral of (1-x)/x^2 can also be solved using the substitution method or integration by parts. However, the formula mentioned in the first question is commonly used and can be derived using these other methods as well.