How do you integrate 1/(e^x(e^x+1))1ex(ex+1) using partial fractions?

1 Answer
Andrea S.
Mar 4, 2017

int dx/(e^x(e^x+1)) = ln(1+e^-x) -e^-x +C∫dxex(ex+1)=ln(1+e−x)−e−x+C

Explanation:

Rather than using partial fractions directly we can prepare the function for a substitution. As the substitution will have to remove the exponential, let's bring it to the numerator noting that: 1/e^x = e^-x1ex=e−x:

int dx/(e^x(e^x+1)) = int (e^-xdx)/((e^x+1))∫dxex(ex+1)=∫e−xdx(ex+1)

Substitute now e^-x = te−x=t, dt =-e^-xdxdt=−e−xdx

int dx/(e^x(e^x+1)) = - int (dt)/(1/t+1) = - int(tdt)/(t+1) ∫dxex(ex+1)=−∫dt1t+1=−∫tdtt+1

Now separate in partial fractions simply adding and subtracting 11 to the numerator:

int dx/(e^x(e^x+1)) = -int (t+1-1)/(t+1)dt = -int dt + int (dt)/(t+1)∫dxex(ex+1)=−∫t+1−1t+1dt=−∫dt+∫dtt+1

This are regular integrals we can solve straight away:

int dx/(e^x(e^x+1)) = -t+ln abs (t+1) +C∫dxex(ex+1)=−t+ln|t+1|+C

and undoing the substitution:

int dx/(e^x(e^x+1)) = ln(1+e^-x) -e^-x +C∫dxex(ex+1)=ln(1+e−x)−e−x+C

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