MHB Help Solving an Integral: I've Tried, But Can't Seem to Get It

[tmt1]

I need to solve this integral. I've had this problem for about a month and I'm still not sure how to solve it:

$$\int_{}^{} \frac{e^x}{e^{2x} + 1}\,dx$$

I've tried a few methods, but I haven't gotten very far.

 

[Prove It]

I need to solve this integral. I've had this problem for about a month and I'm still not sure how to solve it:

$$\int_{}^{} \frac{e^x}{e^{2x} + 1}\,dx$$

I've tried a few methods, but I haven't gotten very far.

Substitute $\displaystyle \begin{align*} \mathrm{e}^x = \tan{ \left( \theta \right) } \implies \mathrm{e}^x\,\mathrm{d}x = \sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{\mathrm{e}^x}{\mathrm{e}^{2\,x} + 1} \,\mathrm{d}x} &= \int{ \frac{\sec^2{ \left( \theta \right) } }{\tan^2{ \left( \theta \right) } + 1 } \,\mathrm{d}\theta } \\ &= \int{ \frac{\sec^2{ \left( \theta \right) } }{ \sec^2{ \left( \theta \right) } }\,\mathrm{d}\theta } \\ &= \int{ 1\,\mathrm{d}\theta } \\ &= \theta + C \\ &= \arctan{ \left( \mathrm{e}^x \right) } + C \end{align*}$

 

[soroban]

$$\int_{}^{} \frac{e^x}{e^{2x} + 1}\,dx$$

\text{Let }\,u = e^x \quad\Rightarrow\quad du \,=\,e^x\,dx

\text{Substitute: }\;\int \frac{du}{u^2+1}

\text{Integrate: }\; \arctan u + C

\text{Back-substitute: }\;\arctan(e^x) + C