When $x$ is used to represent a variable, the sum of one and square of variable $x$ is written as $1+x^2$ and the indefinite integration of the reciprocal of algebraic expression $1+x^2$ is written in the following mathematical form.

$\displaystyle \int{\dfrac{1}{1+x^2}\,\,}dx$

The inverse tangent function is written as $\tan^{-1}{x}$ or $\arctan{(x)}$. Now, let us learn how to prove the integration formula for the reciprocal of sum of one and square of variable in integral calculus.

The inverse tangent function is defined in $x$. Hence, the differentiation of the inverse tan function must have to do with respect to $x$. As per the derivative rule of inverse tan function, the differentiation of the inverse tan function can expressed in a rational expression.

$\dfrac{d}{dx}{\, \tan^{-1}{(x)}}$ $\,=\,$ $\dfrac{1}{1+x^2}$

Now, add an arbitrary constant $c$ to the inverse tan function and then evaluate the differentiation of the expression $\tan^{-1}{(x)}+c$ with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\, \Big(\tan^{-1}{(x)}+c\Big)}$

Now, differentiate the inverse trigonometric expression with respect to $x$ by the sum rule of differentiation.

$\implies$ $\dfrac{d}{dx}{\, \Big(\tan^{-1}{(x)}+c\Big)}$ $\,=\,$ $\dfrac{d}{dx}{\,\tan^{-1}{(x)}}$ $+$ $\dfrac{d}{dx}{\, (c)}$

We know the derivative of the inverse tan function but the derivative of constant is zero as per the derivative rule of constant.

$\implies$ $\dfrac{d}{dx}{\, \Big(\tan^{-1}{(x)}+c\Big)}$ $\,=\,$ $\dfrac{1}{1+x^2}+0$

$\implies$ $\dfrac{d}{dx}{\, \Big(\tan^{-1}{(x)}+c\Big)}$ $\,=\,$ $\dfrac{1}{1+x^2}$

Now, let’s evaluate the indefinite integration of the rational expression in algebraic form. The integral of the rational expression is written in the following mathematical form in integral calculus.

$\displaystyle \int{\dfrac{1}{1+x^2}\,\,}dx$

Here, the primitive or an anti-derivative of rational expression is equal to the sum of inverse tan function and the constant of integration.

$\implies$ $\dfrac{d}{dx}{\, \Big(\tan^{-1}{(x)}+c\Big)}$ $\,=\,$ $\dfrac{1}{1+x^2}$ $\,\Longleftrightarrow\,$ $\displaystyle \int{\dfrac{1}{1+x^2}\,\,}dx$ $\,=\,$ $\tan^{-1}{(x)}+c$

$\therefore \,\,\,\,\,\,$ $\displaystyle \int{\dfrac{1}{1+x^2}\,\,}dx$ $\,=\,$ $\tan^{-1}{(x)}+c$

Therefore, it is proved that the indefinite integral of the multiplicative inverse of sum of one and square of variable is equal to the sum of inverse tan function and constant of integration.

Latest Math Topics

Jul 20, 2023

Jun 26, 2023

Jun 23, 2023

Latest Math Problems

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved