int(e^(2x)+1)/(e^x-1)dxe2x+1ex1dx=?

enter image source here

2 Answers
Apr 14, 2018

int(e^(2x)+1)/(e^x-1)dxe2x+1ex1dx

Taking e^x=uex=u
=> du = e^x dxdu=exdx

int(e^(2x)+1)/(e^x-1)dx =int(u^2+1)/(u-1) (du)/e^x e2x+1ex1dx=u2+1u1duex

=>int(u^2+1)/(u-1) xx(du)/u u2+1u1×duu

=>int(u^2+1)/(u(u-1))du u2+1u(u1)du

=>int(u^2+1)/(u^2-u)du u2+1u2udu

That's option E :)

Apr 14, 2018

int \frac{e^{2x} + 1}{e^x - 1} dx = [ int \frac{u^2 + 1}{u^2 - u} du ]_{u = e^x}e2x+1ex1dx=[u2+1u2udu]u=ex

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 dxe2x+1ex1dx=ex+1exex1exdx.

Written like this, we see that the integrand may be viewed as a composed function f(g(x))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}.