Integral of (3x+4)/(x+1) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 3 x$.

      Then let $du = 3 dx$ and substitute $du$:

      $$\int \frac{u + 4}{u + 3}\, du$$

      1. Let $u = u + 3$.

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

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

         1. Rewrite the integrand:

            $$\frac{u + 1}{u} = 1 + \frac{1}{u}$$

         2. Integrate term-by-term:

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

               $$\int 1\, du = u$$

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

            The result is: $u + \log{\left(u \right)}$

         Now substitute $u$ back in:

         $$u + \log{\left(u + 3 \right)} + 3$$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

         $$\int 3\, dx = 3 x$$

      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:

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

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

   Method #3

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

      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: $3 x - 3 \log{\left(x + 1 \right)} + 4 \log{\left(x + 1 \right)}$

2. Add the constant of integration:

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

The answer is:

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