Math Doubts

Proof of Integral rule for Reciprocal of Linear expression

Let $a$ and $b$ represent constants, and $x$ represents a variable. There are two possible ways to form the linear expressions in one variable.

  1. $ax+b$
  2. $ax-b$

The indefinite integral of the multiplicative inverse of the linear expression in one variable with respect to $x$ is written in calculus as follows.

$\displaystyle \int{\dfrac{1}{ax \pm b} \,}dx$

Now, let us derive the integral rule for the reciprocal of linear expression in one variable mathematically.

Transform Linear expression by differentiation

Let $u = ax\pm b$, then differentiate the equation with respect to $x$.

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{d}{dx}{\, (ax \pm b)}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{d}{dx}{\, (ax)} \pm \dfrac{d}{dx}{\, (b)}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{d}{dx}{\, (a \times x)} \pm 0$

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

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

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

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

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

$\implies$ $\displaystyle \int{\dfrac{1}{ax \pm b} \,}dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{u} \,}\Big(\dfrac{du}{a}\Big)$

Simplify the Indefinite integral expression

In the above step, we have successfully converted the integral of rational expression in terms of $u$ from $x$. Now, let’s simplify this indefinite integral expression.

$\implies$ $\displaystyle \int{\dfrac{1}{u} \,}\Big(\dfrac{du}{a}\Big)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{u} \times \dfrac{1}{a} \times \,}du$

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

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

Evaluate the integration of Rational expression

Now, evaluate the indefinite integral of the reciprocal of a variable $u$ with respect to $u$ by the reciprocal rule of integration.

$\implies$ $\displaystyle \dfrac{1}{a} \times \int{ \dfrac{1}{u} \,}du$ $\,=\,$ $\dfrac{1}{a} \times \Big(\log_e{|u|}+c_1\Big)$

$=\,\,\,$ $\dfrac{1}{a} \times \Big(\log_e{|ax\pm b|}+c_1\Big)$

$=\,\,\,$ $\dfrac{1}{a} \times \log_e{|ax\pm b|}+\dfrac{1}{a} \times c_1$

$=\,\,\,$ $\dfrac{1}{a} \times \log_e{|ax\pm b|}+\dfrac{1 \times c_1}{a}$

$=\,\,\,$ $\dfrac{1}{a} \times \log_e{|ax\pm b|}+\dfrac{c_1}{a}$

$=\,\,\,$ $\dfrac{1}{a}\log_e{|ax\pm b|}+c$

$=\,\,\,$ $\dfrac{1}{a}\ln{|ax\pm b|}+c$

Math Doubts

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

Maths Topics

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

Maths Problems

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

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved