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

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

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

   1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

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

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

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

         1. Let $u = x - 1$.

            Then let $du = dx$ and substitute $du$:

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

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

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

            Now substitute $u$ back in:

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

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

      Method #2

      1. Rewrite the integrand:

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

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

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

         1. Rewrite the integrand:

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

         2. Let $u = x - 1$.

            Then let $du = dx$ and substitute $du$:

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

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

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

            Now substitute $u$ back in:

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

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

      Method #3

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

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

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

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

         1. Let $u = x - 1$.

            Then let $du = dx$ and substitute $du$:

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

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

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

            Now substitute $u$ back in:

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

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

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

2. Add the constant of integration:

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

The answer is:

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