Im having trouble understand how to solve the following:

integral from ln(3/4) to ln(4/3) of (e^(2x))/((1+e^x)^(3/2))

please someone show detailed steps on how to solve this, id really appreciate it!!!!!!!!

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Full Text / QuestionIm having trouble understand how to solve the following: integral from ln(3/4) to ln(4/3) of (e^(2x))/((1+e^x)^(3/2)) please someone show detailed steps on how to solve this, id really appreciate …

Im having trouble understand how to solve the following:

integral from ln(3/4) to ln(4/3) of (e^(2x))/((1+e^x)^(3/2))

please someone show detailed steps on how to solve this, id really appreciate it!!!!!!!!

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ms_pinkfloyd420 says

Trig not necessary here. Notice that on top we have

(e^x)(e^x)dx

which is equal to]

(e^x) d(e^x)

suggesting a substitution of y = e^x

Hence dy = e^x dx

and we have the integral of

[(e^x)dy] / [1 + y)^(3/2)

= ?[y dy / (1 + y)^(3/2)]

Now substitute u = 1+y, so du = dy:

?[(u – 1)du/u^(3/2)]

= ?[u^(-1/2) – u^(-3/2)]du

= [2u^(1/2) + 2u^(-1/2)]

replace u by 1 + e^x and get

2[(1 + e^x)^(1/2) + (1 + e^x)^(-1/2)]

Substitute the given values, using e^(ln(3/4) = 3/4 etc:

2[(1 + 3/4)^(1/2) + (1 + 3/4)^(-1/2)] – 2[(1 + 4/3)^(1/2) + (1 + 4/3)^(-1/2)]

= 2[?7/2 + 2/?7] – 2[?(7/3) + ?(3/7)]

NOTE this one could also be done by parts. With

u = e^x and

v = -2(1 +e^x)^(-1/2)

we have

?u dv

= uv – ?vdu

= -2e^x (1 +e^x)^(-1/2) + 2?(1 +e^x)^(-1/2) d(e^x)

= -2e^x (1 +e^x)^(-1/2) + 4(1 +e^x)^(1/2)

= [-2e^x + 4(1 + e^x)]/?(1 + e^x)

= [4 + 2e^x)]/?(1 + e^x)

= [2 + 2(1+e^x)]/?(1 + e^x)

= 2/?(1 + e^x) + 2?(1 + e^x)