Question

How do you evaluate the definite integral $\int{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}}$ from $\left[ 1,{{e}^{6}} \right]$ ?

Answer 1

We have been given to calculate the definite integral of a fractional expression written in terms of a logarithmic function. It has to be integrated with respect to $dx$. We will perform the integration of this expression by using suitable substitutions. Since we have to perform definite integration, we shall also apply the upper limit and lower limits of integration.

Complete step by step solution:
Given that we have to find definite integral of $\int{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}}$ within the interval $\left[ 1,{{e}^{6}} \right]$.
From this statement, we understand that the upper limit of the definite integral is ${{e}^{6}}$ and the lower limit of definite integral is 1.
Thus, the definite integral is given as
$\int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}}$
In order to integrate this expression, we shall make use of a substitution method.
Let $t=1+\ln \left( x \right)$ …………………… equation (1)
Since differential of $\ln x$ is $\dfrac{1}{x}$ and differential of a constant is equal to zero, thus differentiating both sides, we get
$dt=\dfrac{dx}{x}$
Applying the substitution of variable-t and dt, we get
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}}=\int\limits_{1}^{{{e}^{6}}}{\dfrac{dt}{t}}$
We know that the integral is, $\int{\dfrac{dt}{t}}=\ln t+C$.
Substituting this value, we get
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}}=\left. \ln t \right|_{1}^{{{e}^{6}}}$
Now, we shall re-substitute the value of variable-t from equation (1).
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\left. \ln \left( 1+\ln \left( x \right) \right) \right|_{1}^{{{e}^{6}}}$
Applying the upper and lower limits of integration, we get
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\ln \left( 1+\ln \left( {{e}^{6}} \right) \right)-\ln \left( 1+\ln \left( 1 \right) \right)$
Since, $\ln 1=0$, thus, we get
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\ln \left( 1+\ln \left( {{e}^{6}} \right) \right)-\ln \left( 1+0 \right)$
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\ln \left( 1+\ln \left( {{e}^{6}} \right) \right)-\ln \left( 1 \right)$
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\ln \left( 1+\ln \left( {{e}^{6}} \right) \right)-0$
Here, we will use the property of logarithm, $\ln {{a}^{b}}=b\left( \ln a \right)$.
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\ln \left( 1+6\ln e \right)$
Here, since $\ln e=1$, thus, we get
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\ln \left( 1+6\left( 1 \right) \right)$
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=\ln \left( 7 \right)$
Putting $\ln 7=1.9459$, we have
$\Rightarrow \int\limits_{1}^{{{e}^{6}}}{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}.dx}=1.9459$
Therefore, the definite integral $\int{\dfrac{dx}{x\left( 1+\ln \left( x \right) \right)}}$ from $\left[ 1,{{e}^{6}} \right]$ is equal to 1.9459.

Note:
Definite integral of a function $f\left( x \right)$ is the area bound under the graph of function, $y=f\left( x \right)$and above the x-axis which is bound between two bounds as $x=a$ and $x=b$. Here, $a=$ 1 and $b={{e}^{6}}$. One of the powers of integral calculus is that the one of the two boundaries of the area to be found is a curve and the other boundary is the x-axis when the function is being integrated with respect to $dx$.