Question

How do you integrate $\dfrac{1}{{\ln (x)}}$?

Answer 1

Here in this question, we have to find the integral value of the given function. The given function is of the form of logarithmic function. We should know the differentiation formulas and the integration formulas for some functions. Hence, we can determine the solution for the given question.

Complete step-by-step solution:
The integration is a reciprocal or inverse of the differentiation. In integration we have two kinds namely, definite integral and indefinite integral. In definite integral we have limit points and in indefinite integral we don’t have limit points.
Now consider the given function $\dfrac{1}{{\ln (x)}}$
Now we have to integrate the given function $\int {\dfrac{1}{{\ln (x)}}} \,dx$ ---- (1)
We can’t integrate the function directly so we integrate by substitution.
Let we substitute $\ln (x) = t$---- (2)
Take antilog on both sides we have
$\Rightarrow {e^{\ln (x)}} = {e^t}$
On simplifying we get
$\Rightarrow x = {e^t}$
On differentiating the above equation we have
$\Rightarrow dx = {e^t}\,dt$----- (3)
Let we substitute equation (2) and equation (3) in the equation (1) and this can be written as
$\int {\dfrac{{{e^t}}}{t}} \,dt$
As we know the maclurin’s series expansion for the ${e^t}$ is given as
$\Rightarrow {e^t} = 1 + \dfrac{t}{{1!}} + \dfrac{{{t^2}}}{{2!}} + \dfrac{{{t^3}}}{{3!}} + ...$
When we divide the above inequality by t we have
$\Rightarrow \dfrac{{{e^t}}}{t} = \dfrac{1}{t} + 1 + \dfrac{t}{{2!}} + \dfrac{{{t^2}}}{{3!}} + ...$
On applying the integration the above inequality we get
$\Rightarrow \int {\dfrac{{{e^t}}}{t}dt} = \int {\left( {\dfrac{1}{t} + 1 + \dfrac{t}{{2!}} + \dfrac{{{t^2}}}{{3!}} + ...} \right)} dt$
On integrating we get
$\Rightarrow \int {\dfrac{{{e^t}}}{t}dt} = \ln t + t + \dfrac{{{t^2}}}{{2!.2}} + \dfrac{{{t^3}}}{{3!.3}} + ...$
Resubstituting the value of t we get
$\Rightarrow \int {\dfrac{{{e^t}}}{t}dt} = \ln \,\ln x + \ln x + \dfrac{{{{(\ln x)}^2}}}{{2!.2}} + \dfrac{{{{(\ln x)}^3}}}{{3!.3}} + ...$
Hence we have integrated the given logarithmic function and found the solution for the given question.

Note: We know that we have Taylor’s series expansion and maclurin’s series expansion for some special functions. We should know about the integration formulas and differentiation formulas. In integration, applying the integration directly will make it complex so we use the substitution method.