Integral of e^(2x)/(e^(2x)+5)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = e^{2 x}$.

      Then let $du = 2 e^{2 x} dx$ and substitute $du$:

      $$\int \frac{1}{2 u + 10}\, du$$

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = 2 u + 10$.

            Then let $du = 2 du$ and substitute $\frac{du}{2}$:

            $$\int \frac{1}{2 u}\, du$$

            1. The integral of a constant times a function is the constant times the integral of the function:

               $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

               1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

               So, the result is: $\frac{\log{\left(u \right)}}{2}$

            Now substitute $u$ back in:

            $$\frac{\log{\left(2 u + 10 \right)}}{2}$$

         Method #2

         1. Rewrite the integrand:

            $$\frac{1}{2 u + 10} = \frac{1}{2 \left(u + 5\right)}$$

         2. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{1}{2 \left(u + 5\right)}\, du = \frac{\int \frac{1}{u + 5}\, du}{2}$$

            1. Let $u = u + 5$.

               Then let $du = du$ and substitute $du$:

               $$\int \frac{1}{u}\, du$$

               1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

               Now substitute $u$ back in:

               $$\log{\left(u + 5 \right)}$$

            So, the result is: $\frac{\log{\left(u + 5 \right)}}{2}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(2 e^{2 x} + 10 \right)}}{2}$$

   Method #2

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $du$:

      $$\int \frac{e^{u}}{2 e^{u} + 10}\, du$$

      1. Let $u = 2 e^{u} + 10$.

         Then let $du = 2 e^{u} du$ and substitute $\frac{du}{2}$:

         $$\int \frac{1}{2 u}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            So, the result is: $\frac{\log{\left(u \right)}}{2}$

         Now substitute $u$ back in:

         $$\frac{\log{\left(2 e^{u} + 10 \right)}}{2}$$

      Now substitute $u$ back in:

      $$\frac{\log{\left(2 e^{2 x} + 10 \right)}}{2}$$

   Method #3

   1. Let $u = e^{2 x} + 5$.

      Then let $du = 2 e^{2 x} dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{1}{2 u}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         So, the result is: $\frac{\log{\left(u \right)}}{2}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(e^{2 x} + 5 \right)}}{2}$$

2. Add the constant of integration:

   $$\frac{\log{\left(2 e^{2 x} + 10 \right)}}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(2 e^{2 x} + 10 \right)}}{2}+ \mathrm{constant}$$