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

[Levis2]

Homework Statement

Evaluate the integral;
$$\int$$2^x/(2^x+3) dx

The Attempt at a Solution

Now i start out substituting u=2^x+3
Then i get;
$$\int$$2^x/u dx

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

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

Substituting back into the integral;

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

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

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

 

[Subductionzon]

I would recheck your du/dx calculation.

 

[D H]

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

Your mistake is here:

u=2^x+3
du/dx=2^x/ln(2)

Try doing that derivative again.

 

[dextercioby]

I would teach the OP a trick

$$\int \frac{2^x}{2^{x} +3}{}dx= \int \frac{e^{(\ln 2) x}}{e^{(\ln 2)x} +3} dx =...$$

 

[Levis2]

haha oops :P the derivative should be 2^x*ln(2) .. my mistake :)