What is the solution to this integral?

I'm trying to solve the following integral: $\int \frac{x dx}{\sqrt{ax^{2} + bx + c}}$

with the following conditions: $a < 0$, $b > 0$, $c < 0$, and $\left|2ax + b\right| > \sqrt{b^2 - 4ac}$.

I've tried looking this up on the internet and in books of integrals, but I can't find anything.

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Have you tried completing the square first? –  Guess who it is. Jan 31 '12 at 5:36

Look for trigonometric substitution. Complete the square under the root to get

$$\frac{1}{\sqrt{-a}}\int\frac{x}{\sqrt{\alpha^2-(x-\beta)^2}}dx$$

Then the substitution $(x-\beta)= \alpha \sin(t)$ should work.

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I did try that, but I couldn't get $|2ax+b|>\sqrt{b^2−4ac}$ condition, only $|2ax+b|<\sqrt{b^2−4ac}$. It doesn't matter though, since when I rechecked my calculations, the value for x that didn't satisfy $|2ax+b|<\sqrt{b^2−4ac}$ was wrong, and the right value did (it's the solution to a physics problem and the value was one that can't occur). –  aragilar Feb 1 '12 at 1:29

Put it this way

$$\int {\frac{x}{{\sqrt {a{x^2} + bx + c} }}dx} = \frac{1}{{2a}}\int {\frac{{2ax + b}}{{\sqrt {a{x^2} + bx + c} }}dx} - \frac{b}{{2a}}\int {\frac{{dx}}{{\sqrt {a{x^2} + bx + c} }}}$$

Then

$$\int {\frac{x}{{\sqrt {a{x^2} + bx + c} }}dx} = \frac{1}{a}\sqrt {a{x^2} + bx + c} - \frac{b}{{2a}}\int {\frac{{dx}}{{\sqrt {a{x^2} + bx + c} }}}$$

So let's focus on the last integral.

$$a{x^2} + bx + c = a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} + \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}}} \right]$$

Thus by making $${x + \frac{b}{{2a}}} = u$$

we have a new polynomial to integrate. Consider the cases:

1. $\displaystyle \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}} < 0 = - {k^2}$. So you have:

$$\frac{1}{{\sqrt a }}\int {\frac{{du}}{{\sqrt {{u^2} - {k^2}} }}} = \frac{1}{{\sqrt a }}\cosh^{-1} \frac{u}{k}$$

2. $\displaystyle \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}} = 0$. You get

$$\frac{1}{{\sqrt a }}\int {\frac{{du}}{u}} = \frac{1}{{\sqrt a }}\log u$$

1. $\displaystyle \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}} > 0 = {k^2}$

$$\frac{1}{{\sqrt a }}\int {\frac{{du}}{{\sqrt {{u^2} + {k^2}} }}} = \frac{1}{{\sqrt a }}{\sinh ^{ - 1}}\frac{u}{k}$$

So it all depends on the polynomial.

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Since $a<0$ and assumedly $b^2-4ac>0$ (since the OP mentions $|2ax+b|>\sqrt{b^2-4ac}$, and $>$ is hard to define on $\mathbb{C}$), I would think a trig substitution would be better here. –  robjohn Jun 3 '12 at 0:30