Integral $\int\frac{dy}{(y^2 + ay + b)^{3/2}}$ Need the indefinite integral with $a,b$ constants. This is not a homework question. Need it for a physics problem.
 A: We have $$I =\int \frac {1}{(y^2+ay+b)^{\frac {3}{2}}} dy =8\int \frac {1}{((2y+a)^2+4b-a^2)^{\frac {3}{2}}} dy $$ Substitute $u=2y+a $ and then $u=\sqrt {4b-a^2} \tan v $ to get $$I = \frac {1}{4b-a^2}\int \cos v dv $$ Hope you can take it from here. 
A: Hint 
$\displaystyle y^2+ay+b=y^2+ay+(\frac{a}{2})^2+b-(\frac{a}{2})^2 = (y+\frac{a}{2})^2 + \frac{4b-a^2}{4}$
and try the use of two substitutions: one from y to x, another from x to a trig function.

 $\displaystyle\int{\frac{dy}{(y^2+ay+b)^{3/2}}} = \frac{4(2y+a)}{(4b-a^2)\sqrt{(2y+a)^2+4b-a^2)}} +C$

A: $∫\frac{dy}{(y2+ay+b)^{3/2}} = ∫\frac{dy}{((y + \frac{a}{2})^2+\frac{1}{4}(4b-a^2))^{3/2}} = |\frac{a}{2} + y = u , dy = du| = ∫\frac{du}{(u^2+\frac{1}{4}(4b-a^2))^{3/2}} = | t = \frac{2u}{\sqrt{4b - a^2}} , dt = \frac{2du}{\sqrt{4b-a^2}} = \frac{4}{4b-a^2}∫\frac{dt}{(t^2+1)^{3/2}}= | t = tan(z) , dt = \frac{dz}{cos^2(z)}| = \frac{4}{4b-a^2}∫cos(z)dz = \frac{4sin(z)}{4b-a^2} + C= \frac{4sin(arctan(t))}{4b-a^2}+ C=-\frac{4t}{(a^2-4b)\sqrt{t^2+1}} + C= \frac{8u}{(4b-a^2)^{3/2}\sqrt{\frac{a^2-4b-4u^2}{a^2-4b}}} + C= \frac{2(a+2y)}{(4b-a^2)\sqrt{y^2 + ay+b}} + C $
