Evaluate the integral: I = integral ((1+e^x)²)/(1+e^2x) dx

Evaluate the integral: $$ I = \int \frac{(1+e^x)^2}{1+e^{2x}}\,dx. $$

Step 1 · Expand the numerator

Expanding $(1+e^x)^2=1+2e^x+e^{2x}$, the integral splits into three terms:

$$ I = \int \frac{1}{1+e^{2x}}\,dx + 2\int \frac{e^x}{1+e^{2x}}\,dx + \int \frac{e^{2x}}{1+e^{2x}}\,dx. $$

Step 2 · Simplify the third term

Rewrite the third term:

$$ \frac{e^{2x}}{1+e^{2x}} = 1 - \frac{1}{1+e^{2x}}, $$

so that:

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

Adding this to the first term, the two $\int \frac{1}{1+e^{2x}}\,dx$ pieces cancel exactly, leaving just $x$.

Step 3 · Evaluate the middle term

For $\displaystyle 2\int \frac{e^x}{1+e^{2x}}\,dx$, let $t=e^x$, so $dt=e^x\,dx$:

$$ 2\int \frac{dt}{1+t^2} = 2\arctan t = 2\arctan(e^x). $$

Step 4 · Combine all results

Adding the pieces from Steps 2 and 3:

$$ I = x + 2\arctan(e^x) + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $1+e^{2x}>0$ always.
• The key simplification is recognizing that splitting off a "$1$" from $e^{2x}/(1+e^{2x})$ produces a term that exactly cancels the leftover $1/(1+e^{2x})$ piece from the constant term in the expansion — this avoids ever having to integrate $1/(1+e^{2x})$ directly.
• The result can be checked by differentiating $x+2\arctan(e^x)$: $\;1+\dfrac{2e^x}{1+e^{2x}}\;$, and verifying this equals $\dfrac{(1+e^x)^2}{1+e^{2x}}=\dfrac{1+2e^x+e^{2x}}{1+e^{2x}}=1+\dfrac{2e^x}{1+e^{2x}}$.