Question

What is the integral of $ \dfrac{1}{2x} $ ?

Answer 1

Hint: We first explain the term $ \dfrac{dy}{dx} $ where $ y=f\left( x \right) $ . We then need to integrate the equation\[\int{\dfrac{1}{2x}dx}\] once to find all the solutions of the differential equation. We take one constant for the integration. We get the equation of a logarithmic function.

Complete step by step solution:
We have to find the integral of the equation $ \dfrac{1}{2x} $ . The mathematical form is \[\int{\dfrac{1}{2x}dx}\].
The main function is $ y=f\left( x \right) $ .
We have to find the antiderivative or the integral form of the equation.
We know the integral form of \[\int{\dfrac{1}{x}dx}=\log \left| x \right|+c\].
Constant terms get separated from the integral.
Simplifying the differential form,
We get \[\int{\dfrac{1}{2x}dx}=\dfrac{1}{2}\int{\dfrac{1}{x}dx}=\dfrac{1}{2}\log \left| x \right|+c\].
Here $ c $ is another constant.
The integral form of the equation $ \dfrac{1}{2x} $ is \[\dfrac{1}{2}\log \left| x \right|+c\].
So, the correct answer is “\[\dfrac{1}{2}\log \left| x \right|+c\]”.

Note: The solution of the differential equation is the equation of a logarithmic function. The first order differentiation of \[\dfrac{1}{2}\log \left| x \right|+c\] gives the tangent of the circle for a certain point which is equal to $ \dfrac{dy}{dx}=\dfrac{1}{2x} $ .