
Evaluate the given definite integration: $\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}$
Answer
510.9k+ views
Hint: We will solve this integration with the help of a substitution method. Once we get a simple expression of known integral, we will execute the integration. Then we will apply the given limits to get the definite of the integration of $\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}$.
Complete step-by-step answer:
The integral given to us is $\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}$.
To solve this integral, we will make use of a substitution method.
We will substitute log x = t.
To get the substitution for dx, we will differentiate both sides of log x = t.
$\begin{align}
& \Rightarrow d\left( \log x \right)=dt \\
& \Rightarrow \dfrac{dx}{x}=dt \\
& \Rightarrow dx=xdt \\
\end{align}$
Therefore, we will substitute dx = xdt in the integral.
We will also find the new limits after the substitution.
The lower limit of the integral is $e$.
Thus, we will substitute x = $e$ in log x = t to find the value of t.
$\Rightarrow \log e=t$
But we know that log e = 1
$\Rightarrow $ t = 1.
The upper limit of the integral is ${{e}^{2}}$.
Thus, we will substitute x = ${{e}^{2}}$ in log x = t to find the value of t.
$\Rightarrow \log {{e}^{2}}=t$
But we know that the $\log {{e}^{2}}=2\log e$ and we also know that log e = 1.
$\Rightarrow $ t = 2
Therefore, the new lower limit of the integral is 1 and the new upper limit of the integral is 2.
$\Rightarrow \int\limits_{1}^{2}{\dfrac{xdt}{xt}}$
The x in the numerator and denominator gets divided.
$\Rightarrow \int\limits_{1}^{2}{\dfrac{dt}{t}}$
Now, the integral is a known integral and hence we will now execute the integration.
\[\Rightarrow \int\limits_{1}^{2}{\dfrac{dt}{t}}=\left[ \log t \right]_{1}^{2}\]
Now, we will apply the limits of the integration.
\[\Rightarrow \left[ \log t \right]_{1}^{2}=\log \left( 2 \right)-\log \left( 1 \right)\]
But we know that the log 1 = 0
Hence, the integral $\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}$ gets evaluated as log 2.
Note: It is always beneficial to use substitution methods for solving complex integrals. Students are advised to remember that it is important to change the limits also.
Complete step-by-step answer:
The integral given to us is $\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}$.
To solve this integral, we will make use of a substitution method.
We will substitute log x = t.
To get the substitution for dx, we will differentiate both sides of log x = t.
$\begin{align}
& \Rightarrow d\left( \log x \right)=dt \\
& \Rightarrow \dfrac{dx}{x}=dt \\
& \Rightarrow dx=xdt \\
\end{align}$
Therefore, we will substitute dx = xdt in the integral.
We will also find the new limits after the substitution.
The lower limit of the integral is $e$.
Thus, we will substitute x = $e$ in log x = t to find the value of t.
$\Rightarrow \log e=t$
But we know that log e = 1
$\Rightarrow $ t = 1.
The upper limit of the integral is ${{e}^{2}}$.
Thus, we will substitute x = ${{e}^{2}}$ in log x = t to find the value of t.
$\Rightarrow \log {{e}^{2}}=t$
But we know that the $\log {{e}^{2}}=2\log e$ and we also know that log e = 1.
$\Rightarrow $ t = 2
Therefore, the new lower limit of the integral is 1 and the new upper limit of the integral is 2.
$\Rightarrow \int\limits_{1}^{2}{\dfrac{xdt}{xt}}$
The x in the numerator and denominator gets divided.
$\Rightarrow \int\limits_{1}^{2}{\dfrac{dt}{t}}$
Now, the integral is a known integral and hence we will now execute the integration.
\[\Rightarrow \int\limits_{1}^{2}{\dfrac{dt}{t}}=\left[ \log t \right]_{1}^{2}\]
Now, we will apply the limits of the integration.
\[\Rightarrow \left[ \log t \right]_{1}^{2}=\log \left( 2 \right)-\log \left( 1 \right)\]
But we know that the log 1 = 0
Hence, the integral $\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}$ gets evaluated as log 2.
Note: It is always beneficial to use substitution methods for solving complex integrals. Students are advised to remember that it is important to change the limits also.
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