Integral of (log10(x))/(x^3) dx

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

   Then let $du = \frac{dx}{x}$ and substitute $\frac{du}{\log{\left(10 \right)}}$:

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

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

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

      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 $\operatorname{dv}{\left(u \right)} = e^{- 2 u}$.

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

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

         1. Let $u = - 2 u$.

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

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

            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: $- \frac{e^{u}}{2}$

            Now substitute $u$ back in:

            $$- \frac{e^{- 2 u}}{2}$$

         Now evaluate the sub-integral.

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

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

         1. Let $u = - 2 u$.

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

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

            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: $- \frac{e^{u}}{2}$

            Now substitute $u$ back in:

            $$- \frac{e^{- 2 u}}{2}$$

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

      So, the result is: $\frac{- \frac{u e^{- 2 u}}{2} - \frac{e^{- 2 u}}{4}}{\log{\left(10 \right)}}$

   Now substitute $u$ back in:

   $$\frac{- \frac{\log{\left(x \right)}}{2 x^{2}} - \frac{1}{4 x^{2}}}{\log{\left(10 \right)}}$$

2. Now simplify:

   $$- \frac{2 \log{\left(x \right)} + 1}{4 x^{2} \log{\left(10 \right)}}$$

3. Add the constant of integration:

   $$- \frac{2 \log{\left(x \right)} + 1}{4 x^{2} \log{\left(10 \right)}}+ \mathrm{constant}$$

The answer is:

$$- \frac{2 \log{\left(x \right)} + 1}{4 x^{2} \log{\left(10 \right)}}+ \mathrm{constant}$$