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

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Homework Statement


Evaluate the integral;
\int2x/(2x+3) dx

The Attempt at a Solution


Now i start out substituting u=2x+3
Then i get;
\int2x/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;
\int2x/u (du*ln(2))/2x
This simplifies to;
\intln(2)/u du
Where ln(2) is simply a constant (atleast that's what i think)
so ln(2)\int1/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
 
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I would recheck your du/dx calculation.
 
Levis2 said:
Any idea of what I've done wrong ? :P

Your mistake is here:
Levis2 said:
u=2x+3
du/dx=2x/ln(2)
Try doing that derivative again.
 
I would teach the OP a trick

\int \frac{2^x}{2^{x} +3}{}dx= \int \frac{e^{(\ln 2) x}}{e^{(\ln 2)x} +3} dx =...
 
haha oops :P the derivative should be 2^x*ln(2) .. my mistake :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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