Integral of (x^4)dx/(1-x^5) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 1 - x^{5}$.

      Then let $du = - 5 x^{4} dx$ and substitute $- \frac{du}{5}$:

      $$\int \left(- \frac{1}{5 u}\right)\, du$$

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

         $$\int \frac{1}{u}\, du = - \frac{\int \frac{1}{u}\, du}{5}$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         So, the result is: $- \frac{\log{\left(u \right)}}{5}$

      Now substitute $u$ back in:

      $$- \frac{\log{\left(1 - x^{5} \right)}}{5}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{x^{4}}{1 - x^{5}} = - \frac{4 x^{3} + 3 x^{2} + 2 x + 1}{5 \left(x^{4} + x^{3} + x^{2} + x + 1\right)} - \frac{1}{5 \left(x - 1\right)}$$

   2. Integrate term-by-term:

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

         $$\int \left(- \frac{4 x^{3} + 3 x^{2} + 2 x + 1}{5 \left(x^{4} + x^{3} + x^{2} + x + 1\right)}\right)\, dx = - \frac{\int \frac{4 x^{3} + 3 x^{2} + 2 x + 1}{x^{4} + x^{3} + x^{2} + x + 1}\, dx}{5}$$

         1. Let $u = x^{4} + x^{3} + x^{2} + x + 1$.

            Then let $du = \left(4 x^{3} + 3 x^{2} + 2 x + 1\right) dx$ and substitute $du$:

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

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            Now substitute $u$ back in:

            $$\log{\left(x^{4} + x^{3} + x^{2} + x + 1 \right)}$$

         So, the result is: $- \frac{\log{\left(x^{4} + x^{3} + x^{2} + x + 1 \right)}}{5}$

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

         $$\int \left(- \frac{1}{5 \left(x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{x - 1}\, dx}{5}$$

         1. Let $u = x - 1$.

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

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

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            Now substitute $u$ back in:

            $$\log{\left(x - 1 \right)}$$

         So, the result is: $- \frac{\log{\left(x - 1 \right)}}{5}$

      The result is: $- \frac{\log{\left(x - 1 \right)}}{5} - \frac{\log{\left(x^{4} + x^{3} + x^{2} + x + 1 \right)}}{5}$

   Method #3

   1. Rewrite the integrand:

      $$\frac{x^{4}}{1 - x^{5}} = - \frac{x^{4}}{x^{5} - 1}$$

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

      $$\int \left(- \frac{x^{4}}{x^{5} - 1}\right)\, dx = - \int \frac{x^{4}}{x^{5} - 1}\, dx$$

      1. Let $u = x^{5} - 1$.

         Then let $du = 5 x^{4} dx$ and substitute $\frac{du}{5}$:

         $$\int \frac{1}{5 u}\, du$$

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

            $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{5}$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            So, the result is: $\frac{\log{\left(u \right)}}{5}$

         Now substitute $u$ back in:

         $$\frac{\log{\left(x^{5} - 1 \right)}}{5}$$

      So, the result is: $- \frac{\log{\left(x^{5} - 1 \right)}}{5}$

2. Add the constant of integration:

   $$- \frac{\log{\left(1 - x^{5} \right)}}{5}+ \mathrm{constant}$$

The answer is:

$$- \frac{\log{\left(1 - x^{5} \right)}}{5}+ \mathrm{constant}$$