Evaluate the integral: I = integral (arctan e^x)/(e^x) dx
Evaluate the integral: $$ I = \int \frac{\arctan e^x}{e^x}\,dx. $$
Explanation of Method
Substituting $t=e^x$ converts the integral into $\int \arctan t/t^2\,dt$, resolved by integration by parts followed by a partial fraction decomposition.
- Substitute $t=e^x$.
- Apply integration by parts with $u=\arctan t$, $dv=t^{-2}dt$.
- Decompose the leftover rational integral via partial fractions.
- Substitute back and simplify.
Step 1 · Substitution
Let $t=e^x$, so $dt=e^x\,dx=t\,dx$, giving $dx=dt/t$:
$$ I = \int \frac{\arctan t}{t}\cdot\frac{dt}{t} = \int \frac{\arctan t}{t^2}\,dt. $$
Step 2 · Integration by parts
Let $u=\arctan t$, $dv=t^{-2}dt$, so $du=\dfrac{dt}{1+t^2}$, $v=-\dfrac1t$:
$$ I = -\frac{\arctan t}{t} + \int \frac{dt}{t(1+t^2)}. $$
Step 3 · Partial fractions
Writing $\dfrac{1}{t(1+t^2)}=\dfrac1t-\dfrac{t}{1+t^2}$:
$$ \int \frac{dt}{t(1+t^2)} = \int \left(\frac1t-\frac{t}{1+t^2}\right) dt = \ln|t| - \frac12\ln(1+t^2). $$
Step 4 · Combine and substitute back
$$ I = -\frac{\arctan t}{t} + \ln|t| - \frac12\ln(1+t^2) + C. $$
Substituting back $t=e^x$, and noting $\ln|e^x|=x$:
$$ I = -\frac{\arctan e^x}{e^x} + x - \frac12\ln(1+e^{2x}) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $t=e^x$ |
| $u$ | $\arctan t$ |
| $dv$ | $t^{-2}dt$ |
Additional Notes
- The integrand is defined for all real $x$.
- The partial fraction identity $\tfrac{1}{t(1+t^2)}=\tfrac1t-\tfrac{t}{1+t^2}$ can be verified directly: combining the right side over a common denominator gives $\tfrac{(1+t^2)-t^2}{t(1+t^2)}=\tfrac{1}{t(1+t^2)}$.
- The result can be checked by differentiating $-\tfrac{\arctan e^x}{e^x}+x-\tfrac12\ln(1+e^{2x})$ and confirming it returns $\arctan(e^x)/e^x$.
Result
The solution to the integral is: