Integration of(1+x-1/x)e^(x+1/x)is?

1 Answer
Apr 6, 2018

#I=e^(x+1/x)x+C#

Explanation:

We want to solve

#I=int(1+x-1/x)e^(x+1/x)dx#

Write this as two separate integrals

#I=int(x-1/x)e^(x+1/x)dx+inte^(x+1/x)dx#

Let's call

#color(green)(I_1=int(x-1/x)e^(x+1/x)dx#

Such that

#I=I_1+inte^(x+1/x)dx#

Use integration by parts

#color(blue)(intudv=uv-intvdu#

Let #color(red)(dv=1dx=>v=x# and

And #color(red)(u=e^(x+1/x)=>du=(1-1/x^2)e^(x+1/x)dx#

#I=I_1+e^(x+1/x)x-intx(1-1/x^2)e^(x+1/x)dx#

#color(white)(I)=I_1+e^(x+1/x)x-int(x-1/x)e^(x+1/x)dx#

#color(white)(I)=I_1+e^(x+1/x)x-I_1#

#color(white)(I)=e^(x+1/x)x+C#