# Integrate $\int \frac{dx}{\sqrt{-ax^2 + bx +c}}$

Is there a way to solve this integral?

$$\int \frac{dx}{\sqrt{-ax^2 + bx +c}}, \;\;\;\;\; \forall\; a,b,c \in (0,+\infty) \;in\;R$$

I have tried it by changing of variable and by parts, but no result. I don't know what else to do.

• What are $a,b,c$? Commented Feb 9, 2020 at 16:56
• What do you mean when you say “I have proved it by changing of variable and by parts”? Commented Feb 9, 2020 at 16:56
• They are constants. But I'll check it now. "Change of variable" and "Parts" are two methods to solve integrals. That's what we call it in my country. Maybe "Change of variable" is also known by "Substitution". Commented Feb 9, 2020 at 16:59
• Do you mean tried instead of proved ? [Italian native ? Mind the false friend.]
– user65203
Commented Feb 9, 2020 at 17:02
• Thank you, tried. Commented Feb 9, 2020 at 17:04

$$\int \frac{dx}{\sqrt{-ax^2 + bx +c}} =\int \frac{dx}{\sqrt{\frac{b^2+4ac}{4a}-a(x-\frac b{2a})^2}}=\frac1{\sqrt a}\sin^{-1} \frac{2ax-b}{\sqrt{b^2+4ac}}$$

Hint #1: $$ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})$$

Hint #2: it is possible to find $$\quad p, q \quad$$ such that: $$\quad (x^2+\frac{b}{a}x+\frac{c}{a}) = (x+p)^2 \pm q$$

Hint #3: a substitution $$\quad x+p = t \quad$$ will work...

Begin by completing the square in $$ax^2+bx+c$$. Then you get something like $$\frac{1}{\sqrt{-a}}\int\frac{dx}{\sqrt{\beta^2 - (x-\alpha)^2 }} = {\frac {1}{\sqrt {-a}}\arcsin \left( {\frac {x-\alpha}{\beta}} \right) }$$ or $$\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\alpha)^2 + \beta^2}} ={\frac {\ln \left( x-\alpha+\sqrt {{\alpha}^{2}-2\,x\alpha+{\beta}^{2 }+{x}^{2}} \right) }{\sqrt {a}}}$$ or $$\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\alpha)^2 - \beta^2}} = {\frac {\ln \left( x-\alpha+\sqrt {{\alpha}^{2}-2\,x\alpha-{\beta}^{2 }+{x}^{2}} \right) }{\sqrt {a}}}$$

Hints.

If $$a=0,$$ the integral is what you can do, presumably. So first let $$a\ne 0.$$

Secondly, the quadratic expression $$ax^2+bx+c$$ may be put either as a perfect square, a sum of two squares, or a difference of two squares depending on the sign of the quantity $$\Delta=b^2-4ac,$$ which is called its discriminant.

If you can write $$ax^2+bx+c$$ as $$(px+q)^2$$ for some $$p,q$$ depending on $$a\ne 0,b,c,$$ then again the integral is presumably one you can do. This happens when $$\Delta=0.$$

Or you may be able to write the quadratic expression as $$(px+q)^2-k^2,$$ in that case you may use a suitable trigonometric substitution (usually involving a sine). Or as $$k^2-(px+q)^2,$$ for which you may use a secant substitution.

Or if the expression is of the form $$(px+q)^2+k^2,$$ then we may use a tangent substitution.

Hint Use the substitution $$$$u = \frac{2 a x + b}{\sqrt{|\Delta|}}$$$$

Sorry, there are remaining sign errors in what follows.

Set

$$ax^2+bx+c=(px+q)^2r+|r|,$$

which is achieved by

$$\begin{cases}a=p^2r,\\b=2pqr,\\c=q^2r+|r|.\end{cases}$$

As the sign of $$r$$ must match that of $$a$$, the solution is

$$\begin{cases}r=\dfrac{|b^2-4ac|}{4a},\\p=\sqrt{\dfrac ar},\\q=\dfrac{b}{2pr}.\end{cases}$$

By applying the change of variable

$$t=px+q,$$

this gives $$\int\frac{dx}{\sqrt{ax^2+bx+c}}=\frac1{p}\int\frac{dt}{\sqrt{|r|(1+\text{sgn(r)}t^2)}},$$

the antiderivatives of which are $$\text{arsinh }t$$, $$\arcsin t$$ or do-not-exist, depending on the signs.

A prime candidate for Euler substitution.

\begin{align*} \int \frac{dx}{\sqrt{-ax^2+bx+c}} &= -2 \int \frac{dt}{\frac{b-2\sqrt c\,t}{a+t^2}t+\sqrt c} \frac{a\sqrt c+bt-\sqrt c\,t^2}{(a+t^2)^2} \\ &= -2 \int \frac{dt}{a+t^2} \\ &= -\frac2{\sqrt a} \tan^{-1}\left(\frac t{\sqrt a}\right) + C \\ &= \boxed{-\frac2{\sqrt a} \tan^{-1}\left(\frac{\sqrt{-ax^2+bx+c}-\sqrt c}{\sqrt a\,x}\right) + C} \end{align*}

$$t=\frac{\sqrt{-ax^2+b+c}-\sqrt c}{x} \iff x=\frac{b-2\sqrt c\,t}{a+t^2} \implies dx = -\frac{2a\sqrt c+2bt-2\sqrt c\,t^2}{(a+t^2)^2}\,dt$$