Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
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
.
Forums
Homework Help
Calculus and Beyond Homework Help
Integration that leads to logarithm functions problem
Reply to thread
Message
[QUOTE="Philip Robotic, post: 6050683, member: 601236"] Hi everyone, So I am a high school student and I am learning calculus by myself right now (pretty new to that stuff still). Currently I am working through some problems where integration leads to logarithm functions. While doing one of the exercises I noticed one thing I don't understand. I think I am missing something obvious. Could someone explain this to me please? So the task is to solve an integral: (1)$$\int { \frac { { x }^{ -\frac { 1 }{ 2 } } }{ { 2x }^{ \frac { 1 }{ 2 } } } } dx$$ I reduced it to (2) $$\int { \frac { 1 }{ 2x } } dx$$ And then solved it, no problem here (yet) (3)$$\int { \frac { 1 }{ 2x } } dx\quad =\quad \int { \frac { 1 }{ 2 } \cdot } \frac { 1 }{ x } dx\quad =\quad \frac { 1 }{ 2 } \cdot \int { \frac { 1 }{ x } dx } \quad =\quad \frac { 1 }{ 2 } \ln { \left| x \right| } +C$$ But I thought as well that I can do it a different way round. I could set the nominator to be the derivative of 2x to use the rule: (4)$$\int { \frac { f'(x) }{ f(x) } } dx\quad =\quad \ln { \left| f(x) \right| } +C$$ (I know I used it previously :wink:) So that is what I did first, as 2 is the derivative i set the nominator as 2 and compensated for it with a fraction: (5)$$\int { \frac { 1 }{ 2x } } dx\quad =\quad \int { \frac { 1 }{ 2 } } \cdot \frac { 2 }{ 2x } dx\quad =\quad \frac { 1 }{ 2 } \int { \frac { 2 }{ 2x } } dx$$ I know the integrand can be simplified but I think it should work either way, shouldn't it? Please correct me if I'm wrong. Back to the solving. So now, according to the rule from the eq. (4) I get something weird: (6)$$\frac { 1 }{ 2 } \int { \frac { 2 }{ 2x } } dx\quad =\quad \frac { 1 }{ 2 } \ln { \left| 2x \right| } +C$$ And the result does not equal to the one I got in the eq. (3). Please help, where have I made a mistake? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Integration that leads to logarithm functions problem
Back
Top