Integral of (1-2*x)^(-10) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \left(- \frac{1}{2 u^{10}}\right)\, du$$

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

         $$\int \frac{1}{u^{10}}\, du = - \frac{\int \frac{1}{u^{10}}\, du}{2}$$

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

            $$\int \frac{1}{u^{10}}\, du = - \frac{1}{9 u^{9}}$$

         So, the result is: $\frac{1}{18 u^{9}}$

      Now substitute $u$ back in:

      $$\frac{1}{18 \left(1 - 2 x\right)^{9}}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{1}{\left(1 - 2 x\right)^{10}} = \frac{1}{\left(2 x - 1\right)^{10}}$$

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

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

      $$\int \frac{1}{2 u^{10}}\, du$$

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

         $$\int \frac{1}{u^{10}}\, du = \frac{\int \frac{1}{u^{10}}\, du}{2}$$

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

            $$\int \frac{1}{u^{10}}\, du = - \frac{1}{9 u^{9}}$$

         So, the result is: $- \frac{1}{18 u^{9}}$

      Now substitute $u$ back in:

      $$- \frac{1}{18 \left(2 x - 1\right)^{9}}$$

   Method #3

   1. Rewrite the integrand:

      $$\frac{1}{\left(1 - 2 x\right)^{10}} = \frac{1}{1024 x^{10} - 5120 x^{9} + 11520 x^{8} - 15360 x^{7} + 13440 x^{6} - 8064 x^{5} + 3360 x^{4} - 960 x^{3} + 180 x^{2} - 20 x + 1}$$

   2. Rewrite the integrand:

      $$\frac{1}{1024 x^{10} - 5120 x^{9} + 11520 x^{8} - 15360 x^{7} + 13440 x^{6} - 8064 x^{5} + 3360 x^{4} - 960 x^{3} + 180 x^{2} - 20 x + 1} = \frac{1}{\left(2 x - 1\right)^{10}}$$

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

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

      $$\int \frac{1}{2 u^{10}}\, du$$

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

         $$\int \frac{1}{u^{10}}\, du = \frac{\int \frac{1}{u^{10}}\, du}{2}$$

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

            $$\int \frac{1}{u^{10}}\, du = - \frac{1}{9 u^{9}}$$

         So, the result is: $- \frac{1}{18 u^{9}}$

      Now substitute $u$ back in:

      $$- \frac{1}{18 \left(2 x - 1\right)^{9}}$$

   Method #4

   1. Rewrite the integrand:

      $$\frac{1}{\left(1 - 2 x\right)^{10}} = \frac{1}{1024 x^{10} - 5120 x^{9} + 11520 x^{8} - 15360 x^{7} + 13440 x^{6} - 8064 x^{5} + 3360 x^{4} - 960 x^{3} + 180 x^{2} - 20 x + 1}$$

   2. Rewrite the integrand:

      $$\frac{1}{1024 x^{10} - 5120 x^{9} + 11520 x^{8} - 15360 x^{7} + 13440 x^{6} - 8064 x^{5} + 3360 x^{4} - 960 x^{3} + 180 x^{2} - 20 x + 1} = \frac{1}{\left(2 x - 1\right)^{10}}$$

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

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

      $$\int \frac{1}{2 u^{10}}\, du$$

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

         $$\int \frac{1}{u^{10}}\, du = \frac{\int \frac{1}{u^{10}}\, du}{2}$$

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

            $$\int \frac{1}{u^{10}}\, du = - \frac{1}{9 u^{9}}$$

         So, the result is: $- \frac{1}{18 u^{9}}$

      Now substitute $u$ back in:

      $$- \frac{1}{18 \left(2 x - 1\right)^{9}}$$

2. Now simplify:

   $$- \frac{1}{18 \left(2 x - 1\right)^{9}}$$

3. Add the constant of integration:

   $$- \frac{1}{18 \left(2 x - 1\right)^{9}}+ \mathrm{constant}$$

The answer is:

$$- \frac{1}{18 \left(2 x - 1\right)^{9}}+ \mathrm{constant}$$