Question 1193106
<pre>

{{{int((dx)/(e^(2x)+1))}}}

Notice that {{{e^(2x)+1}}} can be written as the sum of squares:

{{{(e^x)^2+1^2}}} 

Draw a right triangle
Always put the simpler expression on the adjacent side.  Since 1
is simpler than e<sup>x</sup>, we put it on the adjacent side.  
The sign between the squares is + so we put the more complicated 
expression on the opposite side.  [If it had been - we would have 
put the more complicated expression on the hypotenuse.]

{{{drawing(200,125,-1,5,-1,4,
locate(1,.7,theta),
locate(2,0,1), locate(4.1,1.9,e^x),

triangle(0,0,4,0,4,3) )}}}

So {{{e^x/1^"" = opposite/adjacent = tan(theta)}}}

{{{e^x = tan(theta)}}}
Solve for x by taking natural logs of both sides:
{{{x=ln(tan(theta)^"")}}}
We find dx by multiplying the derivative by d&theta; 

{{{dx=sec^2(theta)*matrix(1,2,d,theta)/tan(theta))}}}

Then we use the Pythagorean theorem to label the
missing side, which is the hypotenuse.

{{{drawing(200,125,-1,5,-1,4,
locate(1,.7,theta),
locate(2,0,1), locate(4.1,1.9,e^x),locate(
0,2.6,sqrt(e^(2x)+1)),

triangle(0,0,4,0,4,3) )}}} 

Now we're ready to substitute:

{{{int((dx)/(e^(2x)+1))}}}

We substitute for dx

{{{int((sec^2(theta)*matrix(1,2,d,theta)/tan(theta))/(e^(2x)+1))}}}

Now here's a trick.  Observe that the denominator 
equals the square of the hypotenuse, and since {{{sec(theta)=hypotenuse/adjacent=sqrt((e^(2x)+1))/1^""}}}

the denominator is just sec<sup>2</sup>(&theta;)

{{{int((sec^2(theta)*matrix(1,2,d,theta)/tan(theta))/(sec^2(theta)))}}}

Now the sec<sup>2</sup>(&theta;)'s cancel:

{{{int((cross(sec^2(theta))*matrix(1,2,d,theta)/tan(theta))/(cross(sec^2(theta))))}}}

{{{int(matrix(1,2,d,theta)/tan(theta))}}}

{{{int(cot(theta)matrix(1,2,d,theta))}}}

{{{ln(sin(theta)^"")+C}}}

Now we look back at the triangle and find that

{{{sin(theta)=opposite/hypotenuse = e^x/sqrt(e^(2x)+1)}}}

{{{ln(e^x/sqrt(e^(2x)+1))+C}}}


{{{ln(e^x)-ln(sqrt(e^(2x)+1))+C}}}


{{{x-ln(e^(2x)+1)^("1/2")+C}}}

{{{x-expr(1/2)ln(e^(2x)+1)+C}}}   <-- final answer

Edwin</pre>