Have you considered Completing the Square and then a Trigonometric substitution?
how do you integrate 1/ ( 6+x+ x^2 )^(1/2)
let 6+x+ x^2 = (x+2)^2 (z^2)
then dx = -10z / ( z^2 +1) ^(2) dz
substituting that into the question, i got
integrate -2 / (z^2+1) and hence the answer was -2 arctan ( ((3-x)/ (x+2))^(1/2))
however the answer should have been arcsin ( (2x-1)/5)).
is there something wrong with my substitution?
Now use the substitution
So the integral becomes
Now get this function back in terms of .
P.S. I do not agree with the answer given. If they want the integral to contain arcsin, the integrand should have been of the form , not as yours was.
Are you sure you copied the question down correctly?