How do you integrate 1ex(ex+1) using partial fractions?

1 Answer
Mar 4, 2017

dxex(ex+1)=ln(1+ex)ex+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: 1ex=ex:

dxex(ex+1)=exdx(ex+1)

Substitute now ex=t, dt=exdx

dxex(ex+1)=dt1t+1=tdtt+1

Now separate in partial fractions simply adding and subtracting 1 to the numerator:

dxex(ex+1)=t+11t+1dt=dt+dtt+1

This are regular integrals we can solve straight away:

dxex(ex+1)=t+ln|t+1|+C

and undoing the substitution:

dxex(ex+1)=ln(1+ex)ex+C