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IAGI Journal Integration of Petrophysical Analysis and Elastic Log
30 août 2021 Integration of Petrophysical Analysis and Elastic Log Properties as an Input to Optimize the Development Wells Target in Unique Globigerina ...
We will then discuss complex integration culminating with the The choice of branch is immaterial for many properties of the logarithm
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If this property is not specified or has not a positive value a default of 20 is taken. 3.3 DIProxy Log Configuration. The default log configuration file for
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5 oct. 2019 Limit the Number of Audit Log Files Kept by Integration Server. ... Java classes methods
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recognise integrals which can lead to logarithm functions. is that if we recognise that the function we are trying to integrate is the derivative of.
mc ty inttologs
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TerraTek services for mechanical property profiling provide a facies-type analysis based on integrating log and mechanical property.
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Integration that leadsto logarithm functions
mc-TY-inttologs-2009-1 The derivative oflnxis1x. As a consequence, if we reverse the process, the integral of1xis lnx+c. In this unit we generalise this result and see how a wide variety of integrals result in logarithm functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: recognise integrals in which the numerator is the derivative of the denominator. rewrite integrals in alternative forms so that the numerator becomes the derivative of the denominator. recognise integrals which can lead to logarithm functions.
Contents
1.Introduction 2
2.Some examples 3
www.mathcentre.ac.uk 1c?mathcentre 2009
1. IntroductionWe already know that when we differentiatey= lnxwe finddy
dx=1x. We also know that if we havey= lnf(x)and we differentiate it we finddy dx=f?(x)f(x). The point is that if we recognise that the function we are trying to integrate is the derivative of another function, we can simply reverse the process. So if the function we are trying to integrate is a quotient, and if the numerator is the derivative of the denominator, then the integral will involve a logarithm: ify= lnf(x)so thatdy dx=f?(x)f(x) and, reversing the process, ?f?(x) f(x)dx= ln(f(x)) +c. This procedure works if the functionf(x)is positive, because then we can take its logarithm. What happens if the function is negative? In that case,-f(x)is positive, so that we can take the logarithm of-f(x). Then: ify= ln(-f(x))so thatdy dx=-f?(x)-f(x)=f?(x)f(x) and, reversing the process, ?f?(x) f(x)dx= ln(-f(x)) +c when the function is negative. We can combine both these results by using the modulus function. Then we can use the formula in both cases, or when the function takes both positive and negative values (or when we don"t know).
Key Point
To integrate a quotient when the numerator is the derivativeof the denominator, we use ?f?(x) f(x)dx= ln|f(x)|+c. www.mathcentre.ac.uk 2c?mathcentre 2009
2. Some examplesExampleFind?
tanxdx.
Recall that we can rewritetanxassinx
cosx. Observe that the derivative ofcosxis-sinx, so that the numerator is very nearly the derivative of the denominator. We make it so by rewritingsinx cosx as--sinx cosxand the integral becomes tanxdx=?sinx cosxdx =-?-sinx cosxdx =-ln|cosx|+c. This result can be written in the alternative formln|secx|+c.
Example
Find?x
1 +x2dx.
The derivative of the denominator is2x. Note that the numerator is not quite the derivative of the denominator, but we can make it so by rewritingx
1 +x2as12·2x1 +x2. Then
x
1 +x2dx=12?
2x1 +x2dx
1
2ln|1 +x2|+c.
Example
Find ?1 xln|x|dx.
Remember that the derivative ofln|x|is1
x. So we rewrite the integrand slightly differently: 1 xln|x|=1/xln|x|. Now the numerator is the derivative of the denominator. So 1 xln|x|dx=?1/xln|x|dx = ln|ln|x||+c. www.mathcentre.ac.uk 3c?mathcentre 2009
ExampleFind?xcosx+ sinx
xsinxdx. First of all think about what we would obtain if we differentiated the denominator: let"s do this first. Ify=xsinx, then using the product rule of differentiation, dy dx=xcosx+ sinx. So we see that in the integral we are trying to find, the numerator is the derivative of the denominator. So ?xcosx+ sinx xsinxdx= ln|xsinx|+c.
Integration that leadsto logarithm functions
mc-TY-inttologs-2009-1 The derivative oflnxis1x. As a consequence, if we reverse the process, the integral of1xis lnx+c. In this unit we generalise this result and see how a wide variety of integrals result in logarithm functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: recognise integrals in which the numerator is the derivative of the denominator. rewrite integrals in alternative forms so that the numerator becomes the derivative of the denominator. recognise integrals which can lead to logarithm functions.
Contents
1.Introduction 2
2.Some examples 3
www.mathcentre.ac.uk 1c?mathcentre 2009
1. IntroductionWe already know that when we differentiatey= lnxwe finddy
dx=1x. We also know that if we havey= lnf(x)and we differentiate it we finddy dx=f?(x)f(x). The point is that if we recognise that the function we are trying to integrate is the derivative of another function, we can simply reverse the process. So if the function we are trying to integrate is a quotient, and if the numerator is the derivative of the denominator, then the integral will involve a logarithm: ify= lnf(x)so thatdy dx=f?(x)f(x) and, reversing the process, ?f?(x) f(x)dx= ln(f(x)) +c. This procedure works if the functionf(x)is positive, because then we can take its logarithm. What happens if the function is negative? In that case,-f(x)is positive, so that we can take the logarithm of-f(x). Then: ify= ln(-f(x))so thatdy dx=-f?(x)-f(x)=f?(x)f(x) and, reversing the process, ?f?(x) f(x)dx= ln(-f(x)) +c when the function is negative. We can combine both these results by using the modulus function. Then we can use the formula in both cases, or when the function takes both positive and negative values (or when we don"t know).
Key Point
To integrate a quotient when the numerator is the derivativeof the denominator, we use ?f?(x) f(x)dx= ln|f(x)|+c. www.mathcentre.ac.uk 2c?mathcentre 2009
2. Some examplesExampleFind?
tanxdx.
Recall that we can rewritetanxassinx
cosx. Observe that the derivative ofcosxis-sinx, so that the numerator is very nearly the derivative of the denominator. We make it so by rewritingsinx cosx as--sinx cosxand the integral becomes tanxdx=?sinx cosxdx =-?-sinx cosxdx =-ln|cosx|+c. This result can be written in the alternative formln|secx|+c.
Example
Find?x
1 +x2dx.
The derivative of the denominator is2x. Note that the numerator is not quite the derivative of the denominator, but we can make it so by rewritingx
1 +x2as12·2x1 +x2. Then
x
1 +x2dx=12?
2x1 +x2dx
1
2ln|1 +x2|+c.
Example
Find ?1 xln|x|dx.
Remember that the derivative ofln|x|is1
x. So we rewrite the integrand slightly differently: 1 xln|x|=1/xln|x|. Now the numerator is the derivative of the denominator. So 1 xln|x|dx=?1/xln|x|dx = ln|ln|x||+c. www.mathcentre.ac.uk 3c?mathcentre 2009
ExampleFind?xcosx+ sinx
xsinxdx. First of all think about what we would obtain if we differentiated the denominator: let"s do this first. Ify=xsinx, then using the product rule of differentiation, dy dx=xcosx+ sinx. So we see that in the integral we are trying to find, the numerator is the derivative of the denominator. So ?xcosx+ sinx xsinxdx= ln|xsinx|+c.