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 $\dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}}$ .