Question

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

Answer 1

`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`