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

When $x$ represents a variable and $a$ represents a constant, the multiplicative inverse of sum of their squares is written in the following mathematical form.

$\dfrac{1}{x^2+a^2}$

The indefinite integral of reciprocal of sum of squares with respect to $x$ is expressed mathematically in the following form.

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

The indefinite integration of multiplicative inverse of sum of squares is equal to the sum of product of reciprocal of constant and inverse tan of quotient of variable by constant, and constant of integration.

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

It can also be written in following mathematical form.

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

The indefinite integration formula for the reciprocal of sum of squares can be written in terms of any variables as exampled here.

$(1) \,\,\,$ $\displaystyle \int{\dfrac{1}{g^2+b^2}\,}dg$ $\,=\,$ $\dfrac{1}{b}\tan^{-1}{\Big(\dfrac{g}{b}\Big)}+c$

$(2) \,\,\,$ $\displaystyle \int{\dfrac{1}{m^2+k^2}\,}dm$ $\,=\,$ $\dfrac{1}{k}\arctan{\Big(\dfrac{m}{k}\Big)}+c$

$(3) \,\,\,$ $\displaystyle \int{\dfrac{1}{z^2+q^2}\,}dz$ $\,=\,$ $\dfrac{1}{q}\tan^{-1}{\Big(\dfrac{z}{q}\Big)}+c$

Evaluate $\displaystyle \int{\dfrac{1}{y^2+2^2}\,}dy$

Take $x = y$ and $a = 2$ and substitute them in this formula to evaluate the integration.

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

Learn how to derive the indefinite integration formula for reciprocal of sum of squares.

Latest Math Topics

Aug 31, 2024

Aug 07, 2024

Jul 24, 2024

Dec 13, 2023

Latest Math Problems

Sep 04, 2024

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved