Integral of 7*dx/(3-2*x)^4 dx

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

   $$\int \frac{7}{\left(3 - 2 x\right)^{4}}\, dx = 7 \int \frac{1}{\left(3 - 2 x\right)^{4}}\, dx$$

   1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

         $$\frac{1}{\left(3 - 2 x\right)^{4}} = \frac{1}{\left(2 x - 3\right)^{4}}$$

      2. Let $u = 2 x - 3$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{1}{2 u^{4}}\, du$$

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

            $$\int \frac{1}{u^{4}}\, du = \frac{\int \frac{1}{u^{4}}\, du}{2}$$

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

               $$\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$$

            So, the result is: $- \frac{1}{6 u^{3}}$

         Now substitute $u$ back in:

         $$- \frac{1}{6 \left(2 x - 3\right)^{3}}$$

      Method #2

      1. Rewrite the integrand:

         $$\frac{1}{\left(3 - 2 x\right)^{4}} = \frac{1}{16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81}$$

      2. Rewrite the integrand:

         $$\frac{1}{16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81} = \frac{1}{\left(2 x - 3\right)^{4}}$$

      3. Let $u = 2 x - 3$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{1}{2 u^{4}}\, du$$

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

            $$\int \frac{1}{u^{4}}\, du = \frac{\int \frac{1}{u^{4}}\, du}{2}$$

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

               $$\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$$

            So, the result is: $- \frac{1}{6 u^{3}}$

         Now substitute $u$ back in:

         $$- \frac{1}{6 \left(2 x - 3\right)^{3}}$$

      Method #3

      1. Rewrite the integrand:

         $$\frac{1}{\left(3 - 2 x\right)^{4}} = \frac{1}{16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81}$$

      2. Rewrite the integrand:

         $$\frac{1}{16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81} = \frac{1}{\left(2 x - 3\right)^{4}}$$

      3. Let $u = 2 x - 3$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{1}{2 u^{4}}\, du$$

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

            $$\int \frac{1}{u^{4}}\, du = \frac{\int \frac{1}{u^{4}}\, du}{2}$$

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

               $$\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$$

            So, the result is: $- \frac{1}{6 u^{3}}$

         Now substitute $u$ back in:

         $$- \frac{1}{6 \left(2 x - 3\right)^{3}}$$

   So, the result is: $- \frac{7}{6 \left(2 x - 3\right)^{3}}$

2. Add the constant of integration:

   $$- \frac{7}{6 \left(2 x - 3\right)^{3}}+ \mathrm{constant}$$

The answer is:

$$- \frac{7}{6 \left(2 x - 3\right)^{3}}+ \mathrm{constant}$$