Question

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

Answer 1

Hint: We first explain the term $\dfrac{dy}{dx}$ where $y=f\left( x \right)$. We then need to integrate the equation\[\int{\dfrac{\arctan x}{{{x}^{2}}}dx}\] once to find all the solutions of the differential equation. We know that $\arctan x={{\tan }^{-1}}x$. 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 $\dfrac{\arctan x}{{{x}^{2}}}$. The mathematical
form is \[\int{\dfrac{ta{{n}^{-1}}x}{{{x}^{2}}}dx}\].
The main function is $y=f\left( x \right)$.
We have to find the anti-derivative or the integral form of the equation.
We assume ${{\tan }^{-1}}x=\theta $ which gives $x=\tan \theta $. We differentiate the
equation with respect to $x$.
$\begin{align}
& d\left( x \right)=d\left( \tan \theta \right) \\
& \Rightarrow dx={{\sec }^{2}}\theta d\theta \\
& \Rightarrow dx=\left( 1+{{\tan }^{2}}\theta \right)d\theta \\
\end{align}$
Now we replace the values in the equation of \[\int{\dfrac{ta{{n}^{-1}}x}{{{x}^{2}}}dx}\] and get
\[\int{\dfrac{ta{{n}^{-1}}x}{{{x}^{2}}}dx}=\int{\dfrac{\theta }{{{\tan }^{2}}\theta }{{\sec }^{2}}\theta d\theta }=\int{\theta {{\csc }^{2}}\theta d\theta }\]
We know the integral form of \[\int{{{\csc }^{2}}xdx}=-\cot x+c\] and \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\].
We use the by parts theorem to find the solution of the integral.
Let’s assume $f\left( x \right)=g\left( x \right)h\left( x \right)$. We need to find the integration
of $\int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx}$.
We take $u=g\left( x \right),v=h\left( x \right)$. This gives $\int{f\left( x \right)dx}=\int{uvdx}$.
The theorem of integration by parts gives $\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx}$.
For our integration \[\int{\theta {{\csc }^{2}}\theta d\theta }\], we take $u=\theta ,v={{\csc
}^{2}}\theta $.
Simplifying the differential form, we get
\[\begin{align}
& \int{\theta {{\csc }^{2}}\theta d\theta }=\theta \int{{{\csc }^{2}}\theta d\theta }-\int{\left( \dfrac{d\theta }{d\theta }\int{{{\csc }^{2}}\theta d\theta } \right)d\theta } \\
&\Rightarrow \int{\theta {{\csc }^{2}}\theta d\theta }=-\theta \cot \theta +\int{\left( \cot \theta
\right)d\theta } \\
\end{align}\].
We also know that \[\int{\left( \cot \theta \right)d\theta }=\log \left| \sin \theta \right|+c\].
Here $c$ is another constant. We replace the value of ${{\tan }^{-1}}x=\theta $.
The integral becomes \[-\theta \cot \theta +\log \left| \sin \theta \right|+c=\dfrac{-{{\tan }^{-1}}x}{x}+\log \left| \dfrac{x}{\sqrt{1+{{x}^{2}}}} \right|+c\]
The integral form of the equation $\dfrac{\arctan x}{{{x}^{2}}}$ is \[\dfrac{-{{\tan }^{-
1}}x}{x}+\log \left| \dfrac{x}{\sqrt{1+{{x}^{2}}}} \right|+c\].

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