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

Solution 1:

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...

Solution 2:

$$\int \frac{dx}{\sqrt{-ax^2 + bx +c}} =\int \frac{dx}{\sqrt{d-a(x-\frac b{2a})^2}}=\frac1{\sqrt a}\sin^{-1}\left(\sqrt{\frac ad}(x-\frac b{2a})\right)+C$$

where $d=\frac{b^2}{4a}+c$ and $(\sin^{-1}t) '= \frac1{\sqrt{1-t^2}}$ is used.

Solution 3:

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 $$ \frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\alpha)^2 - \beta^2}} = {\frac {\ln \left( x-\alpha+\sqrt {{\alpha}^{2}-2\,x\alpha-{\beta}^{2 }+{x}^{2}} \right) }{\sqrt {a}}} $$

Solution 4:

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.