Integral of sqrt(5x+1)^2 dx

1. Let $u = \sqrt{5 x + 1}$.

   Then let $du = \frac{5 dx}{2 \sqrt{5 x + 1}}$ and substitute $du$:

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

   1. Integrate term-by-term:

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

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

         1. There are multiple ways to do this integral.

            Method #1

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

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

               $$\int \left(\frac{u}{10} - \frac{1}{10}\right)\, du$$

               1. Integrate term-by-term:

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

                     $$\int \frac{u}{10}\, du = \frac{\int u\, du}{10}$$

                     1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

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

                     So, the result is: $\frac{u^{2}}{20}$

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

                     $$\int \left(- \frac{1}{10}\right)\, du = - \frac{u}{10}$$

                  The result is: $\frac{u^{2}}{20} - \frac{u}{10}$

               Now substitute $u$ back in:

               $$\frac{u^{4}}{20} - \frac{u^{2}}{10}$$

            Method #2

            1. Rewrite the integrand:

               $$u \left(\frac{u^{2}}{5} - \frac{1}{5}\right) = \frac{u^{3}}{5} - \frac{u}{5}$$

            2. Integrate term-by-term:

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

                  $$\int \frac{u^{3}}{5}\, du = \frac{\int u^{3}\, du}{5}$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                     $$\int u^{3}\, du = \frac{u^{4}}{4}$$

                  So, the result is: $\frac{u^{4}}{20}$

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

                  $$\int \left(- \frac{u}{5}\right)\, du = - \frac{\int u\, du}{5}$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

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

                  So, the result is: $- \frac{u^{2}}{10}$

               The result is: $\frac{u^{4}}{20} - \frac{u^{2}}{10}$

         So, the result is: $\frac{u^{4}}{10} - \frac{u^{2}}{5}$

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

         $$\int \frac{2 u}{5}\, du = \frac{2 \int u\, du}{5}$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

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

         So, the result is: $\frac{u^{2}}{5}$

      The result is: $\frac{u^{4}}{10}$

   Now substitute $u$ back in:

   $$\frac{\left(5 x + 1\right)^{2}}{10}$$

2. Now simplify:

   $$\frac{\left(5 x + 1\right)^{2}}{10}$$

3. Add the constant of integration:

   $$\frac{\left(5 x + 1\right)^{2}}{10}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(5 x + 1\right)^{2}}{10}+ \mathrm{constant}$$