Where Does the Constant (1/2) Come From When Integrating 2x/(1+2x)?

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In summary: Integrate[ x/(1 + x), x] = x - 1/2 Log[1 + 2 x] + 1/2 or x - 1/2 Log[1 + 2 x]orx - 1/2 Log[1 + 2 x] + c
  • #1
farleyknight
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Homework Statement



When I plug in

Integrate[ x/(1 + x), x]

into Mathematica, I get

x - Log[1 + x]

which I've been able to recreate on paper. However, when I plug in

Integrate[ 2 x/(1 + 2 x), x]

I get

1/2 + x - 1/2 Log[1 + 2 x]

but on paper I get

x - 1/2 Log[1 + 2 x]

My question is, where does the constant (1/2) come from? Is it me or Mathematica that's wrong?
 
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  • #2
Integrate[ x/(1 + x), x] =

x - 1/2 Log[1 + 2 x] + 1/2
or
x - 1/2 Log[1 + 2 x]
or
x - 1/2 Log[1 + 2 x] + c

What's the difference?
 
  • #3
rootX said:
Integrate[ x/(1 + x), x] =

x - 1/2 Log[1 + 2 x] + 1/2
or
x - 1/2 Log[1 + 2 x]
or
x - 1/2 Log[1 + 2 x] + c

What's the difference?

5 or 10 points on a test :) depending on how a teacher would grade such mistakes. I'm using Mathematica to check my work and these kinds of things make me wonder if my procedure was off or not.
 
  • #4
I remember asking same question like two years ago. I don't think there's any difference. Here's the matlab:

>> syms x;
>> int( 2*x/(1 + 2*x), x)
ans =
x-1/2*log(1+2*x)

But, regardless of what constant you get, in the end there's always "c"
one constant+1/2 = another constant ('c'). The question doesn't have one single answer.

(Also, I don't know why Mathematica is giving 1/2 in the end)
 
  • #5
You guys rely WAAAY too much on machines for simple calculations. Back in the native American reservation, I only had small beads for calculus.

Anyway, you are doing it wrong, Mathematica is right.

Int[2x/(1+2x)] dx
Substitute u=1+2x so that dx=du/2
Notice that u-1=2x

Therefore, the integral is
Int[(u-1)/2u] du
which, by splitting it up, is
Int[1/2] du +Int[-1/2u]
which is
u/2 -1/2 ln(u)+c
Don't forget to put it in terms of x

(1+2x)/2-1/2ln(1+2x)+c

This is probably where you messed up you thought (1+2x)/2 is x.
Well, it is actually,

1/2+x-1/2ln(1+2x)+c
 
  • #6
Pinu7 said:
Anyway, you are doing it wrong, Mathematica is right.

I wasn't thinking about substitution then but here's how he/me were doing (I think):

=int (2x / (1+2x))
= int (1-1/(1+2x))
=x - 1/2.ln(1+2x) + c

I wouldn't call missing 1/2 wrong if you are using partial fractions (not substitution).
 
  • #7
rootX said:
I wasn't thinking about substitution then but here's how he/me were doing (I think):

=int (2x / (1+2x))
= int (1-1/(1+2x))
=x - 1/2.ln(1+2x) + c

I wouldn't call missing 1/2 wrong if you are using partial fractions (not substitution).

*Slaps hand on face* forgot that the +c made the 1/2 completely meaningless. Sorry, that's a habit of mine.

Of course, x - 1/2 Log[1 + 2 x] + c is the simplified solution and your professor may take off some points if you hand in my/Mathematica's answer.
 

1. What is the purpose of the "Integrate" function?

The "Integrate" function is used in mathematics and science to calculate the definite or indefinite integral of a given function. It helps in finding the area under a curve and is an important tool in solving various problems in physics, engineering, and other fields.

2. How do you use the "Integrate" function?

To use the "Integrate" function, you need to provide the function that you want to integrate and the variable with respect to which you want to integrate. In this case, the function is 2x/(1+2x) and the variable is x. The function will then return the integrated form of the function, which in this case is ln(1+2x) + C, where C is the constant of integration.

3. What is the difference between definite and indefinite integration?

Definite integration involves finding the numerical value of the area under a curve between two given points. It gives a specific result. On the other hand, indefinite integration involves finding the general formula for the area under a curve. It includes a constant of integration and does not give a specific result.

4. Can the "Integrate" function solve all types of integrals?

No, the "Integrate" function can only solve integrals that have a closed-form solution. This means that it can solve simple integrals such as polynomials, trigonometric functions, and exponential functions. However, it cannot solve integrals that require advanced techniques such as integration by parts, substitution, or partial fractions.

5. How is the "Integrate" function useful in science?

The "Integrate" function is essential in science as it helps in solving problems related to finding the area under a curve, calculating the work done by a force, determining the position of an object, and much more. It is also used in various mathematical and statistical models to analyze and interpret data and make predictions.

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