Question

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

Answer 1

Hint: We first explain the term $ \dfrac{dy}{dx} $ where $ y=f\left( x \right) $ . We then need to integrate the equation\[\int{\dfrac{x}{1+{{x}^{2}}}dx}\] 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{x}{1+{{x}^{2}}} $ . The mathematical form is \[\int{\dfrac{x}{1+{{x}^{2}}}dx}\].
The main function is $ y=f\left( x \right) $ .
We have to find the antiderivative or the integral form of the equation.
We assume $ 1+{{x}^{2}}=z $ . We differentiate the equation with respect to $ x $ .
 $
   \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)=\dfrac{dz}{dx} \\
  \Rightarrow 2xdx=dz \\
  \Rightarrow xdx=\dfrac{1}{2}dz \;
 $
Now we replace the values in the equation of \[\int{\dfrac{x}{1+{{x}^{2}}}dx}\] and get
\[\int{\dfrac{x}{1+{{x}^{2}}}dx}=\int{\dfrac{1}{z}\left( \dfrac{1}{2}dz \right)}=\dfrac{1}{2}\int{\dfrac{1}{z}dz}\]
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 \[\int{\dfrac{x}{1+{{x}^{2}}}dx}=\dfrac{1}{2}\int{\dfrac{1}{z}dz}=\dfrac{1}{2}\log \left| z \right|+c\].
Here $ c $ is another constant.
We replace the value of $ 1+{{x}^{2}}=z $ .
We get \[\int{\dfrac{x}{1+{{x}^{2}}}dx}=\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+c\]
The integral form of the equation $ \dfrac{x}{1+{{x}^{2}}} $ is \[\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+c\].
So, the correct answer is “\[\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+c\].”.

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