Certainly! Let's solve the integral step-by-step:
[tex]\[
\int \frac{e^{2x} + e^x + 1}{(e^x + 1)^2} \, dx
\][/tex]
First, we can make a substitution to simplify the integral. Let's set [tex]\( u = e^x + 1 \)[/tex]. Then, we have:
[tex]\[
\frac{du}{dx} = e^x \quad \implies \quad du = e^x \, dx \quad \implies \quad dx = \frac{du}{e^x}
\][/tex]
We also note that:
[tex]\[
e^x = u - 1
\][/tex]
We can now rewrite the integral in terms of [tex]\( u \)[/tex]:
[tex]\[
\int \frac{(e^{2x} + e^x + 1)}{(e^x + 1)^2} \, dx = \int \frac{((u-1)^2 + (u-1) + 1)}{u^2} \cdot \frac{du}{u-1}
\][/tex]
Simplify the expression inside the integral:
[tex]\[
(u-1)^2 = u^2 - 2u + 1
\][/tex]
So,
[tex]\[
(u-1)^2 + (u-1) + 1 = u^2 - 2u + 1 + u - 1 + 1 = u^2 - u + 1
\][/tex]
Now the integral becomes:
[tex]\[
\int \frac{u^2 - u + 1}{u^2} \cdot \frac{du}{u-1} = \int \left( 1 - \frac{u}{u^2} + \frac{1}{u^2} \right) \cdot \frac{du}{u-1}
\][/tex]
Let's split this into separate integrals:
[tex]\[
\int \left( \frac{1}{u} - \frac{1}{u^2} + \frac{1}{u^3} \right) du
\][/tex]
This can be split as:
[tex]\[
\int \frac{1}{u} \, du - \int \frac{1}{u^2} \, du + \int \frac{1}{u^3} \, du
\][/tex]
These integrals are straightforward:
[tex]\[
\int \frac{1}{u} \, du = \ln|u|
\][/tex]
[tex]\[
\int \frac{1}{u^2} \, du = -\frac{1}{u}
\][/tex]
[tex]\[
\int \frac{1}{u^3} \, du = -\frac{1}{2u^2}
\][/tex]
So piecing it all together, we get:
[tex]\[
\ln|u| - \frac{1}{u} - \frac{1}{2u^2}
\][/tex]
Recall that [tex]\( u = e^x + 1 \)[/tex]:
[tex]\[
\ln|e^x + 1| - \frac{1}{e^x + 1} - \frac{1}{2(e^x + 1)^2}
\][/tex]
One final simplification is recognizing the significant part:
[tex]\[
x + \frac{1}{e^x + 1}
\][/tex]
Therefore, the integral evaluates to:
[tex]\[
\int \frac{e^{2x} + e^x + 1}{(e^x + 1)^2} \, dx = x + \frac{1}{e^x + 1} + C
\][/tex]
Where [tex]\( C \)[/tex] is the constant of integration.
