Question

The value of $\int {{{\log }_{10}}xdx = }$
A) $x{\log _{10}}x + c$
B) $x({\log _{10}}x + {\log _{10}}e) + c$
C) ${\log _{10}}x + c$
D) $x({\log _{10}}x - {\log _{10}}e) + c$

Answer 1

In the given question we are asked to find the integration. To solve this first of all we see that we are given a log function with base $10$. We will first convert it into natural log by converting base $10$ to base $e$. After that we will solve the question using various formulas and substitutions to get the answer.
Formula used: When we cannot directly integrate any function we do it by integration by parts method using ILATE. Here we consider that log function as the first function ${\rm I}$ and the $1$ as second function ${\rm I}{\rm I}$. After that we apply by integration by parts method as,
$\int {({\rm I} \cdot {\rm I}{\rm I}) = {\rm I}\int {{\rm I}{\rm I}} } - \int {(\dfrac{{d{\rm I}}}{{dx}} \cdot \int {{\rm I}{\rm I}} )}$.

Complete step-by-step solution:
We are given to integrate the function ${\log _{10}}x$ with respect to $x$. We will first convert it into a log function with base $e$ using formula ${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$. So,

$$\int {{{\log }_{10}}xdx = } \int {\dfrac{{{{\log }_e}x}}{{{{\log }_e}10}}} dx \\\ \Rightarrow \int {{{\log }_{10}}xdx = } \dfrac{1}{{{{\log }_e}10}}\int {{{\log }_e}x} dx \\\$$

We know that by integration by parts method using ILATE, we can say that here we will consider ${\log _{10}}x$ as the first function and $1$ as the second function. So, we can solve $\int {{{\log }_e}x} dx$ as
$\int {{{\log }_e}x} dx = \int {1 \cdot {{\log }_e}x} dx$,
We solve ahead as,

$$\Rightarrow \int {{{\log }_{10}}xdx = } (lo{g_e}x)\int {1dx} - \smallint (\dfrac{{dlo{g_e}x}}{{dx}} \cdot \int 1 dx)dx \\\ \Rightarrow \int {{{\log }_{10}}xdx = } \dfrac{1}{{{{\log }_e}10}}[(lo{g_e}x) \cdot x - \smallint (\dfrac{1}{x} \cdot xdx)dx] \\\ \Rightarrow \int {{{\log }_{10}}xdx = } \dfrac{1}{{{{\log }_e}10}}(xlo{g_e}x - x) + c \\\ \Rightarrow \int {{{\log }_{10}}xdx = } x\left( {\dfrac{{{{\log }_e}x}}{{{{\log }_e}10}} - \dfrac{1}{{{{\log }_e}10}}} \right) + c \\\ \Rightarrow \int {{{\log }_{10}}xdx = } x\left( {{{\log }_{10}}x - {{\log }_{10}}e} \right) + C \\\$$

Where, $C = \dfrac{c}{{{{\log }_e}10}}$.
Hence our answer to the given integration is option D).

Note: To solve such a question, it is important to convert the functions into a form where we are comfortable in, if we can, This will help us not only in solving the question easily but also to understand the concepts of integration. Integration by parts method used here should also be used carefully as chances of miscalculating any term is high.