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

Latest Math Topics

Jul 24, 2022

Jul 15, 2022

Latest Math Problems

Sep 30, 2022

Jul 29, 2022

Jul 17, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved