Question

What is the antiderivative of $\ln {{\left( x \right)}^{2}}$?

Answer 1

Hint: First of all simplify the given logarithmic function by using the formula $\ln {{a}^{m}}=m\ln a$ and take the constant 2 out of the integral sign. Now, to calculate the integral of $\ln x$, write it as $\ln x\times 1$ and assume $\ln x$ as function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and 1 as function 2 $\left( {{f}_{2}}\left( x \right) \right)$ and apply the rule of integration by parts given as $\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)}=\left[ {{f}_{1}}\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left[ {{f}_{1}}'\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]dx}$ to get the answer. Here, \[{{f}_{1}}'\left( x \right)=\dfrac{d}{dx}\left( {{f}_{1}}\left( x \right) \right)\] .

Complete step by step solution:
Here we are asked to find the antiderivative of the function $\ln {{\left( x \right)}^{2}}$. In other words we need to integrate this function. First let us simplify this logarithmic function. So using the property of log given as $\ln {{a}^{m}}=m\ln a$ we get,
$\Rightarrow \ln {{\left( x \right)}^{2}}=2\ln x$
Now, let us assume the integral as I, so we have,
$\Rightarrow I=\int{2\ln xdx}$
Since 2 is a constant so it can be taken out of the integral sign. Therefore we have,
$\Rightarrow I=2\int{\ln xdx}$
We don’t have any direct formula for the integral of log function so let us apply the integration by parts rule. To apply integration by parts rule we must have a product of two functions but here we have only one function. So to make it a product of two functions we can write $\ln x=\ln x\times 1$ where 1 can be called an algebraic function ${{x}^{0}}$. Therefore the integral becomes,
$\Rightarrow I=2\int{\ln x\times 1dx}$
Now, according to the ILATE rule we have to assume $\ln x$ as the function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and 1 as function 2 $\left( {{f}_{2}}\left( x \right) \right)$. Here, ILATE stands for
I – Inverse trigonometric function
L – Logarithmic function
A – Algebraic function
T – Trigonometric function
E – Exponential function
The numbering of the functions is done according to the order of appearance in the above list. Therefore assuming $\ln x$ as function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and 1 as function 2 $\left( {{f}_{2}}\left( x \right) \right)$we have the formula $\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)dx}=\left[ {{f}_{1}}\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left[ {{f}_{1}}'\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]dx}$ to calculate the product of two functions. So we get,
\[\begin{align}
  & \Rightarrow I=2\left[ \ln x\times \int{1dx}-\int{\left( \left( \int{1dx} \right)\times \dfrac{d\left( \ln x \right)}{dx} \right)dx} \right] \\
 & \Rightarrow I=2\left[ \ln x\times \int{dx}-\int{\left( \left( \int{dx} \right)\times \dfrac{d\left( \ln x \right)}{dx} \right)dx} \right] \\
\end{align}\]
We know that $\int{dx}=1$ and $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$, so we can simplify the above integral as:
\[\begin{align}
  & \Rightarrow I=2\left[ \ln x\times x-\int{\left( x\times \dfrac{1}{x} \right)dx} \right] \\
 & \Rightarrow I=2\left[ x\ln x-\int{dx} \right] \\
 & \Rightarrow I=2\left[ x\ln x-x \right] \\
 & \Rightarrow I=2x\left[ \ln x-1 \right] \\
\end{align}\]
Hence the above function is our answer.

Note: You may also think of the other way if you don’t know how to do the numbering of functions according to ILATE but you know the formula as integration by parts. See that we don’t have a direct formula to find the integral of $\ln x$ so we cannot assume it as the function 2 whereas we can easily find the integral of 1 therefore it can be considered as the function 2. However, this is a rough idea only so try to remember the ILATE rule.