Integral of (x+4)/(x+1) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\frac{x + 4}{x + 1} = 1 + \frac{3}{x + 1}$$

   2. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

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

         $$\int \frac{3}{x + 1}\, dx = 3 \int \frac{1}{x + 1}\, dx$$

         1. Let $u = x + 1$.

            Then let $du = dx$ 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:

            $$\log{\left(x + 1 \right)}$$

         So, the result is: $3 \log{\left(x + 1 \right)}$

      The result is: $x + 3 \log{\left(x + 1 \right)}$

   Method #2

   1. Rewrite the integrand:

      $$\frac{x + 4}{x + 1} = \frac{x}{x + 1} + \frac{4}{x + 1}$$

   2. Integrate term-by-term:

      1. Rewrite the integrand:

         $$\frac{x}{x + 1} = 1 - \frac{1}{x + 1}$$

      2. Integrate term-by-term:

         1. The integral of a constant is the constant times the variable of integration:

            $$\int 1\, dx = x$$

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

            $$\int \left(- \frac{1}{x + 1}\right)\, dx = - \int \frac{1}{x + 1}\, dx$$

            1. Let $u = x + 1$.

               Then let $du = dx$ 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:

               $$\log{\left(x + 1 \right)}$$

            So, the result is: $- \log{\left(x + 1 \right)}$

         The result is: $x - \log{\left(x + 1 \right)}$

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

         $$\int \frac{4}{x + 1}\, dx = 4 \int \frac{1}{x + 1}\, dx$$

         1. Let $u = x + 1$.

            Then let $du = dx$ 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:

            $$\log{\left(x + 1 \right)}$$

         So, the result is: $4 \log{\left(x + 1 \right)}$

      The result is: $x + 4 \log{\left(x + 1 \right)} - \log{\left(x + 1 \right)}$

2. Add the constant of integration:

   $$x + 3 \log{\left(x + 1 \right)}+ \mathrm{constant}$$

The answer is:

$$x + 3 \log{\left(x + 1 \right)}+ \mathrm{constant}$$