Integral of ln(x)/x^5 d0

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \log{\left(x \right)}$.

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

      $$\int u e^{- 4 u}\, du$$

      1. Use integration by parts:

         $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

         Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{- 4 u}$.

         Then $\operatorname{du}{\left(u \right)} = 1$.

         To find $v{\left(u \right)}$:

         1. Let $u = - 4 u$.

            Then let $du = - 4 du$ and substitute $- \frac{du}{4}$:

            $$\int \left(- \frac{e^{u}}{4}\right)\, du$$

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

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

            Now substitute $u$ back in:

            $$- \frac{e^{- 4 u}}{4}$$

         Now evaluate the sub-integral.

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

         $$\int \left(- \frac{e^{- 4 u}}{4}\right)\, du = - \frac{\int e^{- 4 u}\, du}{4}$$

         1. Let $u = - 4 u$.

            Then let $du = - 4 du$ and substitute $- \frac{du}{4}$:

            $$\int \left(- \frac{e^{u}}{4}\right)\, du$$

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

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

            Now substitute $u$ back in:

            $$- \frac{e^{- 4 u}}{4}$$

         So, the result is: $\frac{e^{- 4 u}}{16}$

      Now substitute $u$ back in:

      $$- \frac{\log{\left(x \right)}}{4 x^{4}} - \frac{1}{16 x^{4}}$$

   Method #2

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{1}{x^{5}}$.

      Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$.

      To find $v{\left(x \right)}$:

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

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

      Now evaluate the sub-integral.

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

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

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

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

      So, the result is: $\frac{1}{16 x^{4}}$

2. Now simplify:

   $$- \frac{4 \log{\left(x \right)} + 1}{16 x^{4}}$$

3. Add the constant of integration:

   $$- \frac{4 \log{\left(x \right)} + 1}{16 x^{4}}+ \mathrm{constant}$$

The answer is:

$$- \frac{4 \log{\left(x \right)} + 1}{16 x^{4}}+ \mathrm{constant}$$