
What is the integral of $\int{\dfrac{\ln x}{x}.dx}$?
Answer
509.4k+ views
Hint: For solving this question you should know about the integration by parts method. This question contains more than one part for integration and so we will divide in more than one part and then we solve them and in the last we will put the values in that and then by solving that we get our answer.
Complete step-by-step answer:
According to our question we have to calculate the integral of $\int{\dfrac{\ln x}{x}.dx}$. As we can see that there are more than one function and we can assume $\ln \left( x \right)=u,v=\ln \left( x \right)$. And if we differentiate both then we will get,
$\dfrac{du}{dx}=\dfrac{1}{x},\dfrac{dv}{dx}=\dfrac{1}{x}$
As we know that the integration by parts is done as,
$\begin{align}
& I=\int{\left( \dfrac{1}{x} \right)\ln \left( x \right).dx} \\
& =\ln \left( x \right)\ln \left( x \right)-\int{\ln \left( x \right)}.\left( \dfrac{1}{x} \right)dx \\
& =\ln \left( x \right)\ln \left( x \right)-I \\
\end{align}$
So, we will take $I$ on the left side of the equation and then we can further write is as,
$\begin{align}
& 2I={{\left( \ln \left( x \right) \right)}^{2}} \\
& I=\dfrac{1}{2}{{\left( \ln \left( x \right) \right)}^{2}}+c \\
\end{align}$
Note: For calculating the integration of any function always see that if that is in any special form or in a normal form and if that is in any special form then always solve that according to that method because that will be easy and accurate for that and if we solve that with our own method then it becomes difficult and it can be wrong also. So, make that in the proper form. And if that is like $f'\left( x \right)f\left( x \right)$ form, then solve that as $\int{u.v.dx}$ method which is integration by parts method.
Complete step-by-step answer:
According to our question we have to calculate the integral of $\int{\dfrac{\ln x}{x}.dx}$. As we can see that there are more than one function and we can assume $\ln \left( x \right)=u,v=\ln \left( x \right)$. And if we differentiate both then we will get,
$\dfrac{du}{dx}=\dfrac{1}{x},\dfrac{dv}{dx}=\dfrac{1}{x}$
As we know that the integration by parts is done as,
$\begin{align}
& I=\int{\left( \dfrac{1}{x} \right)\ln \left( x \right).dx} \\
& =\ln \left( x \right)\ln \left( x \right)-\int{\ln \left( x \right)}.\left( \dfrac{1}{x} \right)dx \\
& =\ln \left( x \right)\ln \left( x \right)-I \\
\end{align}$
So, we will take $I$ on the left side of the equation and then we can further write is as,
$\begin{align}
& 2I={{\left( \ln \left( x \right) \right)}^{2}} \\
& I=\dfrac{1}{2}{{\left( \ln \left( x \right) \right)}^{2}}+c \\
\end{align}$
Note: For calculating the integration of any function always see that if that is in any special form or in a normal form and if that is in any special form then always solve that according to that method because that will be easy and accurate for that and if we solve that with our own method then it becomes difficult and it can be wrong also. So, make that in the proper form. And if that is like $f'\left( x \right)f\left( x \right)$ form, then solve that as $\int{u.v.dx}$ method which is integration by parts method.
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