Question

`int(e^(2x)+1)/(e^x-1)dx`=?

Answer 1

`int(e^(2x)+1)/(e^x-1)dx`

Taking `e^x=u`
`=> du = e^x dx`

`int(e^(2x)+1)/(e^x-1)dx =int(u^2+1)/(u-1) (du)/e^x`

`=>int(u^2+1)/(u-1) xx(du)/u`

`=>int(u^2+1)/(u(u-1))du`

`=>int(u^2+1)/(u^2-u)du`

That's option E :)

Answer 2

`int \frac{e^{2x} + 1}{e^x - 1} dx = [ int \frac{u^2 + 1}{u^2 - u} du ]_{u = e^x}`

Explanation:

First, we can rewrite the integral as follows:

`int \frac{e^{2x} + 1}{e^x - 1} dx = int \frac{e^x + \frac{1}{e^x}}{e^x - 1} e^x dx`.

Written like this, we see that the integrand may be viewed as a composed function `f(g(x))` with the inner function `g(x) = g'(x) = e^x`. Therefore,

`int \frac{e^x + \frac{1}{e^x}}{e^x - 1} e^x dx =`

`= [ int \frac{u + \frac{1}{u}}{u - 1} du ]_{u = e^x} =`

`= [ int \frac{\frac{u^2 + 1}{u}}{u - 1} du ]_{u = e^x} =`

`=[ int \frac{u^2 + 1}{u^2 - u} du ]_{u = e^x}`.