Integral of 1/((1-2*x)^3) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

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

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

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

      1. Let $u = 2 x - 1$.

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

         $$\int \frac{1}{2 u^{3}}\, du$$

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

            $$\int \frac{1}{u^{3}}\, du = \frac{\int \frac{1}{u^{3}}\, du}{2}$$

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

               $$\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}$$

            So, the result is: $- \frac{1}{4 u^{2}}$

         Now substitute $u$ back in:

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

      So, the result is: $\frac{1}{4 \left(2 x - 1\right)^{2}}$

   Method #2

   1. Rewrite the integrand:

      $$\frac{1}{\left(1 - 2 x\right)^{3}} = - \frac{1}{8 x^{3} - 12 x^{2} + 6 x - 1}$$

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

      $$\int \left(- \frac{1}{8 x^{3} - 12 x^{2} + 6 x - 1}\right)\, dx = - \int \frac{1}{8 x^{3} - 12 x^{2} + 6 x - 1}\, dx$$

      1. Rewrite the integrand:

         $$\frac{1}{8 x^{3} - 12 x^{2} + 6 x - 1} = \frac{1}{\left(2 x - 1\right)^{3}}$$

      2. Let $u = 2 x - 1$.

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

         $$\int \frac{1}{2 u^{3}}\, du$$

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

            $$\int \frac{1}{u^{3}}\, du = \frac{\int \frac{1}{u^{3}}\, du}{2}$$

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

               $$\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}$$

            So, the result is: $- \frac{1}{4 u^{2}}$

         Now substitute $u$ back in:

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

      So, the result is: $\frac{1}{4 \left(2 x - 1\right)^{2}}$

   Method #3

   1. Rewrite the integrand:

      $$\frac{1}{\left(1 - 2 x\right)^{3}} = \frac{1}{- 8 x^{3} + 12 x^{2} - 6 x + 1}$$

   2. Rewrite the integrand:

      $$\frac{1}{- 8 x^{3} + 12 x^{2} - 6 x + 1} = - \frac{1}{\left(2 x - 1\right)^{3}}$$

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

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

      1. Let $u = 2 x - 1$.

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

         $$\int \frac{1}{2 u^{3}}\, du$$

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

            $$\int \frac{1}{u^{3}}\, du = \frac{\int \frac{1}{u^{3}}\, du}{2}$$

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

               $$\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}$$

            So, the result is: $- \frac{1}{4 u^{2}}$

         Now substitute $u$ back in:

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

      So, the result is: $\frac{1}{4 \left(2 x - 1\right)^{2}}$

2. Add the constant of integration:

   $$\frac{1}{4 \left(2 x - 1\right)^{2}}+ \mathrm{constant}$$

The answer is:

$$\frac{1}{4 \left(2 x - 1\right)^{2}}+ \mathrm{constant}$$