Integral of 1/(100+2t) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 t + 100$.

      Then let $du = 2 dt$ 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:

      $$\frac{\log{\left(2 t + 100 \right)}}{2}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{1}{2 t + 100} = \frac{1}{2 \left(t + 50\right)}$$

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

      $$\int \frac{1}{2 \left(t + 50\right)}\, dt = \frac{\int \frac{1}{t + 50}\, dt}{2}$$

      1. Let $u = t + 50$.

         Then let $du = dt$ 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:

         $$\log{\left(t + 50 \right)}$$

      So, the result is: $\frac{\log{\left(t + 50 \right)}}{2}$

2. Add the constant of integration:

   $$\frac{\log{\left(2 t + 100 \right)}}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(2 t + 100 \right)}}{2}+ \mathrm{constant}$$