Complete the square at the denominator:
int dx/(5-4x-x^2)^(5/2) = int dx/(9- (2+x)^2)^(5/2)∫dx(5−4x−x2)52=∫dx(9−(2+x)2)52
int dx/(5-4x-x^2)^(5/2) = 1/3^5 int dx/(1- ((2+x)/3)^2)^(5/2)∫dx(5−4x−x2)52=135∫dx(1−(2+x3)2)52
Subtitute:
(2+x)/3 = sint2+x3=sint
with t in (-pi/2,pi/2)t∈(−π2,π2)
so that in the interval costcost is positive and:
x = -2+3sintx=−2+3sint
dx = 3cost dtdx=3costdt
Then:
int dx/(5-4x-x^2)^(5/2) = 1/3^4 int (costdt)/(1- sin^2t)^(5/2)∫dx(5−4x−x2)52=134∫costdt(1−sin2t)52
and as:
(1-sin^2t)^(5/2) = (sqrt(1-sin^2t))^5 = cos^5t(1−sin2t)52=(√1−sin2t)5=cos5t
we have:
int dx/(5-4x-x^2)^(5/2) = 1/3^4 int dt/cos^4t = 1/3^4 int sec^4t dt∫dx(5−4x−x2)52=134∫dtcos4t=134∫sec4tdt
Solve the resulting integral using the identity sec^2t = 1+tan^2tsec2t=1+tan2t:
int sec^4t dt = int sec^2t * sec^2t dt∫sec4tdt=∫sec2t⋅sec2tdt
int sec^4t dt = int sec^2t (1+tan^2t) dt∫sec4tdt=∫sec2t(1+tan2t)dt
int sec^4t dt = int sec^2t dt +int tan^2tsec^2t dt∫sec4tdt=∫sec2tdt+∫tan2tsec2tdt
int sec^4t dt = tant +tan^3t/3+C∫sec4tdt=tant+tan3t3+C
To undo the substitution note that:
tant = sint/costtant=sintcost
and as we noted that in the interval costcost is positive:
tant = sint/sqrt(1-sin^2t)tant=sint√1−sin2t
so:
tant = ((2+x)/3)/sqrt((1-((2+x)/3)^2)tant=2+x3√(1−(2+x3)2)
tant = (2+x)/(sqrt(9-(2+x)^2)tant=2+x√9−(2+x)2
tant = (2+x)/sqrt(5-4x-x^2)tant=2+x√5−4x−x2
Then:
int dx/(5-4x-x^2)^(5/2) = 1/3^4 ((2+x)/sqrt(5-4x-x^2) + (2+x)^3/(3(5-4x-x^2)^(3/2)))∫dx(5−4x−x2)52=134⎛⎝2+x√5−4x−x2+(2+x)33(5−4x−x2)32⎞⎠
and simplifying:
int dx/(5-4x-x^2)^(5/2) = (2+x)/(81sqrt(5-4x-x^2)) (1+ (2+x)^2/(3(5-4x-x^2)))∫dx(5−4x−x2)52=2+x81√5−4x−x2(1+(2+x)23(5−4x−x2))
int dx/(5-4x-x^2)^(5/2) = (2+x)/(81sqrt(5-4x-x^2)) ((15-12x-3x^2 + 4+4x+x^2)/(3(5-4x-x^2)))∫dx(5−4x−x2)52=2+x81√5−4x−x2(15−12x−3x2+4+4x+x23(5−4x−x2))
int dx/(5-4x-x^2)^(5/2) = ((2+x)(19-8x-2x^2))/(243(5-4x-x^2)^(3/2))∫dx(5−4x−x2)52=(2+x)(19−8x−2x2)243(5−4x−x2)32
