I need to evaluate $\int \frac{dx}{-ax^2 + b}$ while both $a$ and $b$ are positive.

I was blocked while I was trying $ x=\tan\theta $ which turned $ dx=\sec^2\theta\, d\theta $

This method didn't actually work because $-ax^2$ is negative so the substitution didn't return $\int d\theta$ like I was hoping it would.

I'd appreciate any methods or hints!

  • $\begingroup$ $d\tanh\theta = (1 \color{red}{-} \tanh^2\theta)d\theta$. $\endgroup$ Aug 31, 2014 at 9:02
  • $\begingroup$ TeX tips: use x^2 for $x^2$, \tan for $\tan$, and similarly for other common functions. $\endgroup$
    – user147263
    Aug 31, 2014 at 15:22

3 Answers 3


Case $\#1:$ If $a=0,$ we can solve it easily

Case $\#2:$ else

$$-ax^2+b=-a\left(x^2-\dfrac ba\right)$$

Case $\#2A:$ If $\dfrac ba>0, \dfrac ba=c^2$(say)

Write $\dfrac{2c}{x^2-c^2}=\dfrac1{x-c}-\dfrac1{x+c}$

Case $\#2B:$ If $\dfrac ba<0, \dfrac ba=-c^2$(say)

$\int\dfrac{dx}{x^2+c^2}=\dfrac1c\arctan \dfrac xc+K$

Case $\#2C:$ If $b=0$

  • $\begingroup$ I didn't get why do we have $2c$ in the numerator. Case #2.4 $\endgroup$
    – rand
    Aug 31, 2014 at 9:06
  • $\begingroup$ @rand, $$\dfrac1{x-c}-\dfrac1{x+c}=?$$ $\endgroup$ Aug 31, 2014 at 9:07
  • $\begingroup$ I didn't start with a $2c$ in the numerator, to get it I'd need to multiply and devide by $2c$ then take the $1/2c$ out of the integral. $\endgroup$
    – rand
    Aug 31, 2014 at 9:17
  • 2
    $\begingroup$ I found a particule moving faster than the light. Guess $who$ ! $\endgroup$ Aug 31, 2014 at 9:20
  • 1
    $\begingroup$ @ClaudeLeibovici, Who OR What? $\endgroup$ Aug 31, 2014 at 9:41

Which can be obtained:either by putting $x=\sqrt{b/a}\sin \theta$ $$I=\int \frac{dx}{-ax^2 + b}=\int\frac{\sqrt{b/a}\cos \theta d\theta }{b\cos^2\theta}=\sqrt{\frac{a}{b}}\int\sec\theta d\theta=\sqrt{\frac{a}{b}}(\ln|\sec\theta+\tan\theta|)+C$$ Since $\sin\theta=x\sqrt{a/b},\cos\theta=\sqrt{1-x^2(a/b)}$ So: $$I=\sqrt{a/b}\ln\left(\left|\frac1{\sqrt{1-x^2(a/b)}}+\frac{x\sqrt{a/b}}{\sqrt{1-x^2(a/b)}}\right|\right)$$ $$I=\sqrt{a/b}\left[\ln\left|1+x\sqrt{a/b}\right|-\ln\left|\sqrt{1-x^2(a/b)}\right|\right]$$


Since $a$ and $b$ are positive, you can take their principal square roots. So, simple partial fraction expansion solves the problem easily. For some $\alpha$ and $\beta$:

$$\frac{\alpha}{\sqrt{b} - \sqrt{a}x} + \frac{\beta}{\sqrt{b} + \sqrt{a} x} = \frac{1}{b - ax^2}$$

You could do the partial fraction expansion, and then integrate each term. However, this isn't quite the way I'd do it, because it introduces a bunch of fiddle factors which are always tricky to manipulate.

Instead, I'd use the fact that an expansion of this form exists, and conclude that the integral must be of the form:

$$\int \frac{dx}{b - ax^2} = A \log (\sqrt{b} + \sqrt{a}x) + B \log (\sqrt{b} - \sqrt{a}x) + C$$

for some constants $A$ and $B$, and then work backwards. Differentiating both sides gives:

$$\frac{1}{b - ax^2} = \frac{A \sqrt{a}}{\sqrt{b} + \sqrt{a}x} - \frac{B\sqrt{a}}{\sqrt{b} - \sqrt{a}x}$$


$$1 = A \sqrt{a} (\sqrt{b} - \sqrt{a}x) - B\sqrt{a}(\sqrt{b} + \sqrt{a}x)$$ $$= (A - B) \sqrt{ab} - (A + B) a x$$

Matching up coefficients:

$$\sqrt{ab}(A-B) = 1$$ $$a(A+B) = 0$$

This is a system of linear equations, two equations in two unknowns. Because $a$ and $b$ are nonzero, there is a unique solution:

$$A = \frac{1}{2\sqrt{ab}}$$ $$B = -\frac{1}{2\sqrt{ab}}$$

That is:

$$\int \frac{dx}{b - ax^2} = \frac{1}{2\sqrt{ab}} \left( \log (\sqrt{b} - \sqrt{a}x) - \log (\sqrt{b} + \sqrt{a}x) \right) + C$$

You can now use your favourite logarithm identities to rearrange this if you care. For example, if you like hyperbolic trig, you could use:

$$\tanh^{-1} x = \frac{1}{2} \log \left( \frac{1-x}{1+x} \right)$$

to get:

$$\int \frac{dx}{b - ax^2} = \frac{1}{\sqrt{ab}} \tanh^{-1} \left( \sqrt{\frac{a}{b}} x \right) + C$$


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