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Question

Question: Evaluate the given definite integration: \(\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}\)...

Evaluate the given definite integration: ee2dxxlogx\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}

Explanation

Solution

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 ee2dxxlogx\int\limits_{e}^{{{e}^{2}}}{\dfrac{dx}{x\log x}}.

Complete step-by-step answer :
The integral given to us is ee2dxxlogx\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.
d(logx)=dt dxx=dt dx=xdt \begin{aligned} & \Rightarrow d\left( \log x \right)=dt \\\ & \Rightarrow \dfrac{dx}{x}=dt \\\ & \Rightarrow dx=xdt \\\ \end{aligned}
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 ee.
Thus, we will substitute x = ee in log x = t to find the value of t.
loge=t\Rightarrow \log e=t
But we know that log e = 1
\Rightarrow t = 1.
The upper limit of the integral is e2{{e}^{2}}.
Thus, we will substitute x = e2{{e}^{2}} in log x = t to find the value of t.
loge2=t\Rightarrow \log {{e}^{2}}=t
But we know that the loge2=2loge\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.
12xdtxt\Rightarrow \int\limits_{1}^{2}{\dfrac{xdt}{xt}}
The x in the numerator and denominator gets divided.
12dtt\Rightarrow \int\limits_{1}^{2}{\dfrac{dt}{t}}
Now, the integral is a known integral and hence we will now execute the integration.
12dtt=[logt]12\Rightarrow \int\limits_{1}^{2}{\dfrac{dt}{t}}=\left[ \log t \right]_{1}^{2}
Now, we will apply the limits of the integration.
[logt]12=log(2)log(1)\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 ee2dxxlogx\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.