Let the literals $x$ and $a$ form a rational expression in sum of their squares.

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

In this expression, $a$ represents a constant and $x$ represents a variable. Hence, the indefinite integration of this rational expression should have to do with respect to $x$ and it is written in the following mathematical form in integral calculus.

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

Now, let us evaluate it for using this standard integral function form as an integral rule for reciprocal of sum of squares.

In calculus, there is no direct integral formula for evaluating the integration of the multiplicative inverse of sum of square terms. However, we have an integral rule, which helps us to evaluate the integration of the reciprocal of sum of one and square of a variable. Hence, it is a good idea to convert the reciprocal of sum of squares into sum of one and square of a term.

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

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

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

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

$=\,\,\,$ $\require{cancel} \displaystyle \int{\dfrac{1}{a^2 \times \Big(\dfrac{x^2}{a^2}+\dfrac{\cancel{a^2}}{\cancel{a^2}}\Big)}\,\,}dx$

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

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

$=\,\,\,$ $\displaystyle \int{\Bigg(\dfrac{1}{a^2} \times \dfrac{1}{\dfrac{x^2}{a^2}+1}\Bigg)\,\,}dx$

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

The quotient of squares can be simplfied by the power of a quotient rule.

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

The rational expression is successfully transformed into reciprocal of sum of one and square of a term but we cannot perform the indefinite integration at this moment due to involvement of constant $a$ inside the square of term. So, we have to eliminate the influence of the constant $a$ from the square of term.

Assume, $u \,=\, \dfrac{x}{a}$

Now, differentiate the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\,(u)} \,=\, \dfrac{d}{dx}{\,\Big(\dfrac{x}{a}\Big)}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{d}{dx}{\,\Big(\dfrac{1}{a} \times x\Big)}$

The constant factor can be separated from the differentiation by the constant multiple rule of differentiation.

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{a} \times \dfrac{d}{dx}{\,(x)}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{a} \times \dfrac{dx}{dx}$

Now, use the derivative rule of variable for differentiating the variable with respect to same variable.

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{a} \times 1$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{a}$

$\implies$ $a \times du \,=\, 1 \times dx$

$\implies$ $adu \,=\, dx$

$\implies$ $dx \,=\, adu$

Now, express the integral function in terms of $u$ from $x$.

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

$=\,\,\,$ $\dfrac{1}{a^2} \times \displaystyle \int{\dfrac{1 \times a}{u^2+1}\,\,}du$

$=\,\,\,$ $\dfrac{1}{a^2} \times \displaystyle \int{\dfrac{1 \times a}{1+u^2}\,\,}du$

The constant $a$ can be taken out from the integral operation by the constant multiple formula of integration.

$=\,\,\,$ $\dfrac{1}{a^2} \times a \times \displaystyle \int{\dfrac{1}{1+u^2}\,\,}du$

$=\,\,\,$ $\dfrac{1 \times a}{a^2} \times \displaystyle \int{\dfrac{1}{1+u^2}\,\,}du$

$=\,\,\,$ $\dfrac{a}{a^2} \times \displaystyle \int{\dfrac{1}{1+u^2}\,\,}du$

$=\,\,\,$ $\dfrac{\cancel{a}}{\cancel{a^2}} \times \displaystyle \int{\dfrac{1}{1+u^2}\,\,}du$

$=\,\,\,$ $\dfrac{1}{a} \times \displaystyle \int{\dfrac{1}{1+u^2}\,\,}du$

The indefinite integration of the rational expression can be evaluated by the integral rule of reciprocal of sum of one and square of variable.

$=\,\,\,$ $\dfrac{1}{a} \times \Big(\tan^{-1}{u}+k\Big)$

In this case, the literal $k$ represents the constant of integration. The above expression can be simplified further by the distributive property of multiplication over addition.

$=\,\,\,$ $\dfrac{1}{a} \times \tan^{-1}{u}+\dfrac{1}{a} \times k$

$=\,\,\,$ $\dfrac{1}{a}\tan^{-1}{u}+\dfrac{1 \times k}{a}$

$=\,\,\,$ $\dfrac{1}{a}\tan^{-1}{u}+\dfrac{k}{a}$

$=\,\,\,$ $\dfrac{1}{a}\tan^{-1}{\Big(\dfrac{x}{a}\Big)}+\dfrac{k}{a}$

In the right hand side expression, the second term is a constant, which can be simplify denoted by a constant $c$.

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

Latest Math Topics

Jan 06, 2023

Jan 03, 2023

Jan 01, 2023

Dec 26, 2022

Dec 08, 2022

Latest Math Problems

Jan 31, 2023

Nov 25, 2022

Nov 02, 2022

Oct 26, 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