Question

Solve the following integration: $\int {{{\left( {\log x} \right)}^2}dx}$
A) $x\left[ {{{\left( {\log x} \right)}^2} - 2\log x + 2} \right]$
B) $x\left[ {{{\left( {\log x} \right)}^2} - 2\log x + 1} \right]$
C) $x\left[ {{{\left( {\log x} \right)}^2} - 2\log x - 2} \right]$
D) $x\left[ {{{\left( {\log x} \right)}^2} + 2\log x - 2} \right]$

Answer 1

Here we will first form the integral equation in such a way that we can apply the integration by parts. Then we will use the formula of the integration by parts and perform the integration to get the integration of the required equation.

Complete step by step solution:
The function is $\int {{{\left( {\log x} \right)}^2}dx}$.
We can write the above equation as
$I = \int {\left( {{{\left( {\log x} \right)}^2} \cdot 1} \right)dx}$
Now we will use the formula of the integration by parts i.e. $\int {\left( {u \times v} \right)dx} = u\int {vdx} - \int {\left( {u'\int {vdx} } \right)dx}$ . Therefore by applying this we get
$\Rightarrow I = \int {\left( {{{\left( {\log x} \right)}^2} \cdot 1} \right)dx} = {\left( {\log x} \right)^2}\int {1dx} - \int {\left( {\left( {\dfrac{d}{{dx}}{{\left( {\log x} \right)}^2}} \right)\int {1dx} } \right)dx}$
Now by using the simple integration we will get the simplified equation. Therefore, we get
$\Rightarrow I = {\left( {\log x} \right)^2} \cdot x - \int {\left( {2\log x \times \dfrac{1}{x} \cdot x} \right)dx}$
Multiplying the terms in the bracket, we get
$\Rightarrow I = {\left( {\log x} \right)^2} \cdot x - 2\int {\left( {\log x} \right)dx}$
Now we will similarly find the integration of the $\log x$ by using the formula of the integration by parts. Therefore, we get
$\Rightarrow I = {\left( {\log x} \right)^2} \cdot x - 2\int {\left( {\log x \cdot 1} \right)dx}$
$\Rightarrow I = {\left( {\log x} \right)^2} \cdot x - 2\left( {\log x\int {1dx} - \int {\left( {\dfrac{d}{{dx}}\left( {\log x} \right)\int {1dx} } \right)dx} } \right)$
Integrating the terms, we get
$\Rightarrow I = {\left( {\log x} \right)^2} \cdot x - 2\left( {\log x \cdot x - \int {\left( {\dfrac{1}{x} \cdot x} \right)dx} } \right)$
$\Rightarrow I = x{\left( {\log x} \right)^2} - 2\left( {x\log x - \int {1dx} } \right)$
$\Rightarrow I = x{\left( {\log x} \right)^2} - 2\left( {x\log x - x} \right)$
Now we will open the brackets in the equation. Therefore, we get
$\Rightarrow I = x{\left( {\log x} \right)^2} - 2x\log x + 2x$
Now we will take $x$ common from all the terms of the equation. Therefore, we get
$\Rightarrow I = x\left[ {{{\left( {\log x} \right)}^2} - 2\log x + 2} \right]$
Hence the value of the given integration $\int {{{\left( {\log x} \right)}^2}dx}$ is equal to $x\left[ {{{\left( {\log x} \right)}^2} - 2\log x + 2} \right]$.

So, option A is the correct option.

Note:
Integration is defined as the summation of all the discrete data. Differentiation is the opposite of integration i.e. differentiation of the integration is equal to the value of the function or vice versa. Integration is also known as antiderivative. Here, we can make a mistake by just integrating the logarithmic function without using the product rule. This will give us the wrong answer. We need to remember the different integration formulas to simplify the expression.