Evaluate $\displaystyle \int{\log_{e}{(x)}}\,dx$

According to the logarithms, the natural logarithm of a variable $x$ is mathematically written as $\ln{(x)}$ or $\log_{e}{(x)}$. Hence, the indefinite integration of the natural logarithmic function with respect to $x$ can be written in one of the following forms in the integral calculus.

$(1).\,\,\,$ $\displaystyle \int{\ln{(x)}}\,dx$

$(2).\,\,\,$ $\displaystyle \int{\log_{e}{(x)}}\,dx$

There is no integral rule for the natural logarithmic function in the integral calculus. Hence, it is not possible to find the indefinite integration of the natural logarithmic function directly by an integral rule. Now, we have to think about an alternative method.

$\displaystyle \int{\log_{e}{(x)}}\,dx$

The natural logarithmic function $\ln{x}$ and the differential element $dx$ are multiplying in the integration. Hence, we can try the integration by parts rule in this integration problem.

$\displaystyle \int{u}\,dv$ $\,=\,$ $\displaystyle \int{\log_{e}{(x)}}\,dx$

Find the Derivative of Natural Logarithmic function

Assume $u \,=\, \log_{e}{x}$

Let’s differentiate the natural logarithm of $x$ with respect to $x$ to find the derivative of logarithm. It can done by the derivative rule of logarithms.

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{d}{dx}{\,\Big(\log_{e}{x}\Big)}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{x}$

$\,\,\,\therefore\,\,\,\,\,\,$ $du \,=\, \dfrac{1}{x} \times dx$

Evaluate the Integration of the differential element

Similarly, assume $dv \,=\, dx$

Now, integrate both sides of the equation to find the value of $v$ in $x$.

$\implies$ $\displaystyle \int{}dv \,=\, \displaystyle \int{}dx$

$\implies$ $v+c \,=\, x+c$

$\implies \require{cancel}$ $v+\cancel{c} \,=\, x+\cancel{c}$

$\,\,\,\therefore\,\,\,\,\,\,$ $v \,=\, x$

Evaluate Integration of Natural Logarithmic function

In this step, let’s substitute the values in the integration by parts formula for finding the indefinite integration of the natural logarithm in integral calculus.

$\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $\Big(\log_{e}{x}\Big) \times x$ $\,-\,$ $\displaystyle \int{\Big(x \times \dfrac{1}{x}\Big)}\,dx$

$\implies$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x\log_{e}{x}$ $\,-\,$ $\displaystyle \int{\dfrac{x \times 1}{x}}\,dx$

$\implies$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x\log_{e}{x}$ $\,-\,$ $\displaystyle \int{\dfrac{x}{x}}\,dx$

$\implies$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x\log_{e}{x}$ $\,-\,$ $\displaystyle \int{\dfrac{\cancel{x}}{\cancel{x}}}\,dx$

$\implies$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x\log_{e}{x}$ $\,-\,$ $\displaystyle \int{1}\,dx$

Now, use the integral rule of one for finishing the process of the integration.

$\implies$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x\log_{e}{x}$ $\,-\,$ $(x+c_1)$

$\implies$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x\log_{e}{x}-x-c_1$

$\implies$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x\log_{e}{x}-x+c$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \int{\ln{(x)}}\,dx$ $\,=\,$ $x(\log_{e}{x}-1)+c$

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