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

- $ax+b$
- $ax-b$

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.

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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