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

When $x$ is considered to represent a variable, the sum of one and square of variable $x$ is written as $1+x^2$ mathematically. The inverse tangent function written as $\tan^{-1}{x}$ or $\arctan{(x)}$ in mathematics. The integral of the reciprocal of the expression $1+x^2$ is expressed in the following mathematical form.

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

The indefinite integral of the rational expression with respect to $x$ is equal to the tan inverse of $x$.

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

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

The integral law of reciprocal sum of one and square of variable can be expressed in terms of any variable.

$(1)\,\,\,$ $\displaystyle \int{\dfrac{1}{1+l^2}\,\,}dl$ $\,=\,$ $\tan^{-1}{l}+c$

$(2)\,\,\,$ $\displaystyle \int{\dfrac{1}{1+q^2}\,\,}dq$ $\,=\,$ $\arctan{(q)}+c$

$(3)\,\,\,$ $\displaystyle \int{\dfrac{1}{1+y^2}\,\,}dy$ $\,=\,$ $\tan^{-1}{y}+c$

Learn how to prove the integration formula for the multiplicative inverse of one plus variable squared in integral calculus.

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