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A simple integral - I don't agree with Maple.

  • Thread starter Levis2
  • Start date
  • #1
43
0

Homework Statement


Evaluate the integral;
[tex]\int[/tex]2x/(2x+3) dx

The Attempt at a Solution


Now i start out substituting u=2x+3
Then i get;
[tex]\int[/tex]2x/u dx

Now i express dx by du;
u=2x+3
du/dx=2x/ln(2)
(du*ln(2))/2x=dx

This expression of dx is inserted into my integral, and the 2^x's cancel out;
[tex]\int[/tex]2x/u (du*ln(2))/2x
This simplifies to;
[tex]\int[/tex]ln(2)/u du
Where ln(2) is simply a constant (atleast thats what i think)
so ln(2)[tex]\int[/tex]1/u
And the integral becomes;
ln(2)*ln(u)

Substituting back into the integral;

ln(2)*ln(2x+3)

Now maple didn't give me this result. Instead it gave me the following;
ln(2x+3)/ln(2)

Any idea of what i've done wrong ? :P
 

Answers and Replies

  • #2
I would recheck your du/dx calculation.
 
  • #3
D H
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Any idea of what i've done wrong ? :P
Your mistake is here:
u=2x+3
du/dx=2x/ln(2)
Try doing that derivative again.
 
  • #4
dextercioby
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I would teach the OP a trick

[tex] \int \frac{2^x}{2^{x} +3}{}dx= \int \frac{e^{(\ln 2) x}}{e^{(\ln 2)x} +3} dx =... [/tex]
 
  • #5
43
0
haha oops :P the derivative should be 2^x*ln(2) .. my mistake :)
 

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