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# How do you integrate $\dfrac{{{\left( lnx \right)}^{2}}}{{{x}^{2}}}$?

Last updated date: 19th Jun 2024
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Hint: The integration is the process of finding the antiderivative of a function. It is a similar way to add the slices to make it whole.
For example:
The derivative of $x$ is $1$ and the integration of $1$ i.e. $\int{1}$ is $x$.

Complete step by step solution:Given that $\dfrac{{{\left( lnx \right)}^{2}}}{{{x}^{2}}}$ so we have to integrate this term we can write it as $I=\int{\dfrac{{{\left( lnx \right)}^{2}}}{{{x}^{2}}}\,dx}$
The formula used in this solution is:
$\int{udv}=u\int{dv-\int{\left[ \dfrac{du}{dx}\int{dv} \right]}}$
It is called integration by parts. This formula is used when two functions are in the product and we have to integrate it. The function which is in the product needs to be of the same variable either it will be considered as constant.
For example:
$\int{\sin x\centerdot {{x}^{2}}dx}$
Here $\sin x$ is a function of $x$ and also ${{x}^{2}}$ is a function of $x$.
So, we have,
$I=\int{\dfrac{{{\left( lnx \right)}^{2}}}{{{x}^{2}}}\,dx}\,.......(A)$
Let $z=\ln x$ then differentiating it w.r.t $x$, we get
$\dfrac{dz}{dx}=\dfrac{1}{x}$
$dz=\dfrac{dx}{x}\,.........(1)$
And if $z=lnx$ then ${{e}^{z}}=x\,........(2)$
Putting values of $(1)$ and $(2)$ in equation $(A)$
$I=\int{{{z}^{2}}}\centerdot {{e}^{-z}}dz$
Here, ${{z}^{2}}$ and ${{e}^{-z}}$ are in product so we will apply integration by parts considering,
$u={{z}^{2}}$ so $du=2\centerdot zdz$
$\dfrac{du}{dz}=2z$
$dv={{e}^{-z}}dz$ so integration it we get $v=-{{e}^{-z}}$
$I=\int{u\centerdot dv}=u\int{dv-\int{\left[ \dfrac{du}{dz}\centerdot \int{dv} \right]}}$
$=u\centerdot v-\int{\dfrac{du}{dz}\centerdot v}$
$=-{{z}^{2}}-{{e}^{-z}}-2\int{z\left( -{{e}^{-z}} \right)}dz$
$I=-{{z}^{2}}{{e}^{-z}}+2\int{z{{e}^{-z}}}dz$
Again applying integration by parts on
$\int{z{{e}^{-z}}}dz$
$u=z$ so $v=-{{e}^{-z}}$
$\therefore \,I=z{{e}^{-z}}+2\left( -z{{e}^{-z}}-\int{-{{e}^{-z}}dz} \right)$
$=z{{e}^{-z}}-2z{{e}^{-z}}+2\int{{{e}^{-z}}dz}$
$I=z{{e}^{-z}}-2z{{e}^{-z}}-z{{e}^{-z}}+C$
The question is given in the terms of $x$. So, we need to find the answer in terms of $x$.
If we remember that $z=ln(x)$ we’ll have
$I=-{{e}^{ln(x)}}\left( {{\left( lnx \right)}^{2}}+2lnx+2 \right)+C$
$I=C-\dfrac{{{\left( lnx \right)}^{2}}+2lnx+2}{x}$

In the solution it is not necessary that you should only consider $z$ for solving by parts or $u,v$ and you can take any variable you wish. Keep in mind the formula of integration by parts and you are good to go.
In the terms of ${{\left( lnx \right)}^{2}}$ you might get confused by and may be you end up solving it wrong by considering it as $ln{{x}^{2}}$ this term has square power only on $x$ but in question $lnx$ this whole term is under square power.
The term $ln\left( x \right)=z$ when changes to ${{e}^{z}}=x$ this is because there is a property of logarithm that $\log _{b}^{a}=x$ we can write it as ${{b}^{x}}=a$.