Evaluate the integral of 1 / (a+bx2)

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

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

Stumbling Robot

Evaluate the integral of 1 / (a+bx²)

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