Math Doubts

Integral Rule of Reciprocal of Sum of Squares


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


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.


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.

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

Alternative forms

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.

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