Integral of 3sqrt(5x-7)^2 dx

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

   $$\int 3 \left(\sqrt{5 x - 7}\right)^{2}\, dx = 3 \int \left(\sqrt{5 x - 7}\right)^{2}\, dx$$

   1. Let $u = \sqrt{5 x - 7}$.

      Then let $du = \frac{5 dx}{2 \sqrt{5 x - 7}}$ and substitute $du$:

      $$\int \left(2 u \left(\frac{u^{2}}{5} + \frac{7}{5}\right) - \frac{14 u}{5}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int 2 u \left(\frac{u^{2}}{5} + \frac{7}{5}\right)\, du = 2 \int u \left(\frac{u^{2}}{5} + \frac{7}{5}\right)\, du$$

            1. There are multiple ways to do this integral.

               Method #1

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

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

                  $$\int \left(\frac{u}{10} + \frac{7}{10}\right)\, du$$

                  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:

                        $$\int \frac{7}{10}\, du = \frac{7 u}{10}$$

                     The result is: $\frac{u^{2}}{20} + \frac{7 u}{10}$

                  Now substitute $u$ back in:

                  $$\frac{u^{4}}{20} + \frac{7 u^{2}}{10}$$

               Method #2

               1. Rewrite the integrand:

                  $$u \left(\frac{u^{2}}{5} + \frac{7}{5}\right) = \frac{u^{3}}{5} + \frac{7 u}{5}$$

               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:

                     $$\int \frac{7 u}{5}\, du = \frac{7 \int u\, du}{5}$$

                     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: $\frac{7 u^{2}}{10}$

                  The result is: $\frac{u^{4}}{20} + \frac{7 u^{2}}{10}$

            So, the result is: $\frac{u^{4}}{10} + \frac{7 u^{2}}{5}$

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

            $$\int \left(- \frac{14 u}{5}\right)\, du = - \frac{14 \int u\, du}{5}$$

            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: $- \frac{7 u^{2}}{5}$

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

      Now substitute $u$ back in:

      $$\frac{\left(5 x - 7\right)^{2}}{10}$$

   So, the result is: $\frac{3 \left(5 x - 7\right)^{2}}{10}$

2. Now simplify:

   $$\frac{3 \left(5 x - 7\right)^{2}}{10}$$

3. Add the constant of integration:

   $$\frac{3 \left(5 x - 7\right)^{2}}{10}+ \mathrm{constant}$$

The answer is:

$$\frac{3 \left(5 x - 7\right)^{2}}{10}+ \mathrm{constant}$$