Can you square root an equation for u-sub to simplify down

Question

can i take the square root of this calculus equation to simplify my u-sub in the expression -(1/2)du = (e^-x)^2 dx in other words, can i take the sq root of both sides to get (e^-x)dx by itself apart from (e^-x)^2 dx (on the right side)?

 

 

btw initial problem is taking the def integral from -inf to pos inf of (e^-x)/ (1+ e^-2x) dx  

 

∫_-inf^{inf _{(e^-x)/ (1+ e^-2x) dx}}

 

 

shortly after posting this, i wolframed alpha and realized that if instead you let u be equal to the numerator instead of the entire denominator that you end up being left with a u sub that properly integrates to arctan, but i am still curious about whether the other option is mathematically correct, basically idk what you do then when you end up with the square root on the left side which is the square root of a neg number.. which shouldnt exist.. right?

Answer 1

In both ways you will get the same solution.

 

u=1+e^-2x 

du=-2e-^2xdx i.e dx=-2/(u-1)

 

e^-x=(u-1)^1/2

e^-x dx =-du/(2*(u-1)^1/2)

 

So the function becomes  -du /(2u(u-1)^1/2)

 

solution is -arctan((u-1)^-1/2)

 

Substitute u=1+e-2x 

 

gives -arctan(e^-x)

 

The same answer !!!!!!!!!!!!!