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\begin{aligned} \int \frac{\mathrm{e}^{x}}{9+\mathrm{e}^{2 x}} \mathrm{dx} &=\int \frac{1}{9+\left(\mathrm{e}^{x}\right)^{2}}\left(\mathrm{e}^{x} \mathrm{~d} \mathrm{x}\right) \\ &=\int \frac{1}{3^{2}+u^{2}} \mathrm{du}=\frac{1}{3} \arctan \left(\frac{u}{3}\right)+\mathrm{C}=\frac{1}{3} \arctan \left(\frac{\mathrm{e}^{\mathrm{x}}}{3}\right)+\mathrm{C} . \end{aligned}