Substitution for $\int \frac {dx} {ax^2 + bx + c}$ I'm looking for the substitution that makes easier to solve integral containing quadratic polynomial in denominator (!) when such polynomial cannot be broken into parts (if it can, then it's possible to use partial fraction decomposition). Example:
$$\int \frac {dx} {5x^2 + x - 2}$$
Formulas suggested by wikipedia are hard to remember. So I hope there is some kind of substitution.
link: http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_.2F_.28a_x2_.2B_b_x_.2B_c.29n
 A: HINT:
$$\frac1{ax^2+bx+c}=\frac{4a}{(2ax+b)^2+4ca-b^2}$$
Case $1:$ If $4ca-b^2=0,$ $$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2} $$ Put $y=2ax+b$
Case $2:$ If $4ca-b^2>0, 4ca-b^2=d^2$(say)
$$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2+d^2} $$ 
Put $2ax+b=d\tan\theta$ as $\tan^2\theta+1=\sec^2\theta$
Case $3:$  If $4ca-b^2<0, 4ca-b^2=-d^2$(say)
$$\int \frac{dx}{ax^2+bx+c}=4a\int\frac{dx}{(2ax+b)^2-d^2} $$ 
Put $2ax+b=d\sec\theta$  as $\sec^2\theta-1=\tan^2\theta$
A: An idea:
$$5x^2+x-2=5\left(x-\frac{1-\sqrt{41}}{10}\right)\left(x-\frac{1+\sqrt{41}}{10}\right)$$
and now do partial fractions...
A: If you reject use of complex numbers, and your denominator cannot be factored over the reals, then the method to use is: complete the square in the denominator, which will then suggest a trigonometric substitution.   
Even if it CAN be factored, you could still complete the square and use a hyperbolic substitution...
A: Hint: you can always find the complex roots $x_1$ and $x_2$ of a degree $2$ polynomial 
$$ax^2+bx+c. $$ 
In order to evaluate the integral
$$\int \frac{dx}{(x-x_1)(x-x_2)} $$
try to use the residue theorem on the complex valued function $f(z)=\frac{1}{(z-x-1)(z-x_2)}$
and an appropriate contour integral technique.The result would follow.
