0
$\begingroup$

Need the indefinite integral with $a,b$ constants. This is not a homework question. Need it for a physics problem.

  • 0
    try this here i will hope this will help you https://en.wikipedia.org/wiki/Euler_substitution – 2017-01-22
  • 0
    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – 2017-01-22
  • 2
    Completing the square is a nice first thing to try in cases like this. – 2017-01-22

3 Answers 3

1

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.

1

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.

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

  • 0
    Keep in mind that you can Edit your posts rather than posting a new Answer, even after self-deleting a post. In fact it might be the best approach to self-delete, Edit, and then undelete. – 2017-01-22
  • 0
    @hardmath I'll be sure to keep that in mind, cheers – 2017-01-23
0

$∫\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 $