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.

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