Question

integral of 1/(a-bx)dx​

Answer 1

Answer:

Evaluate the following integral for ab \neq 0:

\[ \int \frac{dx}{a+bx^2}. \]

Since ab \neq 0 we know a \neq 0. Then we write,

\[ \int \frac{dx}{a+bx^2} = \frac{1}{a} \int \frac{dx}{1+\frac{b}{a} x}. \]

Now we consider two cases, ab > 0 and ab < 0. First, the case ab > 0. Then, \frac{b}{a} > 0 (since a and \frac{1}{a} must have the same sign by this exercise, Section I.3.5, Exercise #4). Therefore, \sqrt{\frac{b}{a}} exists in \mathbb{R}. Thus,

\begin{align*} \int \frac{dx}{a+bx^2} &= \frac{1}{a} \int \frac{dx}{1+\frac{b}{a}x^2} \\[9pt] &= \frac{1}{a} \int \frac{dx}{1+ \left( \sqrt{\frac{b}{a}} x \right)^2}. \end{align*}

Now, make the substitution u = \sqrt{\frac{b}{a}} x and du = \sqrt{\frac{b}{a}} \, dx. This gives us,

\begin{align*} \int \frac{dx}{a+bx^2} &= \frac{1}{a} \int \frac{dx}{1+\left( \sqrt{\frac{b}{a}} x \right)^2} \\[9pt] &= \frac{1}{\sqrt{ab}} \int \frac{du}{1+u^2} \\[9pt] &= \frac{1}{\sqrt{ab}} \arctan u + C \\[9pt] &= \frac{1}{\sqrt{ab}} \arctan \left( \sqrt{\frac{b}{a}} x \right) + C. \end{align*}

Now, for the case ab < 0 we have \frac{b}{a} < 0 and so -\frac{b}{a} > 0. Therefore,

\begin{align*} \int \frac{dx}{a+bx^2} &= \frac{1}{a} \int \frac{dx}{1 + \frac{b}{a}x^2} \\[9pt] &= \frac{1}{a} \int \frac{dx}{1 - \left(-\frac{b}{a} \right) x^2} \\[9pt] &= \frac{1}{a} \int \frac{dx}{1 - \left( \sqrt{-\frac{b}{a}} x \right)^2}. \end{align*}

Again, we make a substitution, this time let u = \sqrt{-\frac{b}{a}} x and so du = \sqrt{-\frac{b}{a}} \, dx. Therefore, we have dx = \sqrt{-\frac{a}{b}} du, and so

\begin{align*} \int \frac{dx}{a+bx^2} &= \frac{1}{a} \int \frac{dx}{1 - \left(\sqrt{-\frac{b}{a}} x \right)^2} \\ &= \frac{1}{a} \cdot \sqrt{-\frac{a}{b}} \int \frac{du}{1-u^2} \\ &= \frac{1}{a} \cdot \sqrt{\frac{a^2}{-ab}} \int \frac{du}{1-u^2} \\ &= \frac{a}{|a|\sqrt{-ab}} \int \frac{du}{1-u^2} \\ &= \frac{a}{|a|\sqrt{-ab}} \int \frac{1}{2} \left( \frac{1}{1+u} + \frac{1}{1-u} \right) \, du \\ &= \frac{a}{2|a|\sqrt{-ab}} \left( \int \frac{du}{1+u} + \int \frac{du}{1-u} \right) \\ &= \frac{a}{2|a|\sqrt{-ab}} \left( \log |1+u| - \log |1-u| \right) + C. \end{align*}

Now, we simplify and replace u with \sqrt{-\frac{b}{a}} \, x to obtain

\begin{align*} &= \frac{a}{2|a|\sqrt{-ab}} \left( \log \left|\frac{1+u}{1-u}\right| \right) + C \\ &= \frac{a}{2|a| \sqrt{-ab}} \left( \log \left| \frac{1 + \sqrt{-\frac{b}{a}}\, x}{1 - \sqrt{-\frac{b}{a}} \, x} \right| \right) + C \\ &= \frac{a}{2|a|\sqrt{-ab}} \log \left| \frac{\sqrt{|a|} + x \sqrt{|b|}}{\sqrt{|a|} - x\sqrt{|b|}} \right| \right) + C. \end{align*}