Math Help - Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

1. Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

I'm working my way through a series of ever tougher integrals. I'm stuck at no. 283:

$\int \frac {dx} {x \sqrt {a x^2 + b x + c} }$

... and there are some further even tougher ones.

I understand that it is supposed to evaluate to:

$- \frac 1 {\sqrt c} \ln \left({\frac {2 \sqrt c \sqrt {a x^2 + b x + c} + bx + 2c} x} }\right)$

or:

$- \frac 1 {\sqrt c} \sinh^{-1} \left({ \frac {bx + 2 c} {|x| \sqrt {4 a c - b^2} } }\right)$

depending on the sign of $b^2 - 4 a c$

a) completing the square on the expression under the square root
b) substituting $z = 2 a x + b$ and $z = (2 a x + b)^2$
c) integrating by parts with $u = \sqrt {a x^2 + b x + c}$ and $dv = x$ etc.

I've also tried working backward, differentiating the $\sinh^{-1}$ expression, to see where it gets me, but the penny is not dropping.

Any hints?

The good cause this is for is ProofWiki which now has over 10000 proofs up.

Thanks, guys.

2. Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

It's not elegant but ...

$x \sqrt {a x^2 + b x + c}=\frac{x}{(a x^2 + b x + c)^{-\frac 12}}=\frac{x[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{(a x^2 + b x + c)^{-\frac 12} [2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}=$

$\frac{x[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{[2 \sqrt c +(bx+2c) (ax^2+bx+c)^{-\frac 12}]}=\frac{num}{den}$

where:

$num=\frac{[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{x}$

$den=\frac{[2 \sqrt c +(bx+2c) (ax^2+bx+c)^{-\frac 12}]}{x^2}$

differentiating num w.r.t. x gives:

$num'=\frac{x[\sqrt c (2ax+b)(ax^2+bx+c)^{-\frac 12}+b]-[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{x^2}$

$num'=(ax^2+bx+c)^{-\frac 12} \frac{[\sqrt c (2ax^2+bx)+bx \sqrt{ax^2+bx+c}]-[2 \sqrt c (ax^2+bx+c)+(bx+2c)\sqrt{ax^2+bx+c}])}{x^2}$

$num'=(ax^2+bx+c)^{-\frac 12} \frac{[\sqrt c (2ax^2+bx -2ax^2 -2bx -2c)+bx \sqrt{ax^2+bx+c}]-[(bx+2c)\sqrt{ax^2+bx+c}])}{x^2}$

$num'=(ax^2+bx+c)^{-\frac 12} \frac{[\sqrt c (-bx -2c)-[(2c)\sqrt{ax^2+bx+c}])}{x^2}$

$num'=-\frac{2c +\sqrt c (bx +2c)(ax^2+bx+c)^{-\frac 12}}{x^2}=-\sqrt c \times den$

Now provided num is real and non zero the original integral is

$\int \frac{den}{num}dx=\frac {-1}{\sqrt c}\int \frac {num'}{num}=\frac {-1}{\sqrt c} ln(num)=\frac {-1}{\sqrt c} ln \left(\frac{[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{x} \right)$

3. Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

That works for me. Many thanks.

4. Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

I've not followed it all the way through, but the substitution x = 1/u , followed by a completing of the square and then a second substitution seems to work well.

5. Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

Originally Posted by BobP
I've not followed it all the way through, but the substitution x = 1/u , followed by a completing of the square and then a second substitution seems to work well.
I'm just about to try it now -- thank you for this, it seems to clear the $x$ from the denominator leaving one with the integral of $\frac 1 {\sqrt {a + bx + c x^2} }$ whose solution I have already evaluated.