Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 26

intx^2/(3+4x-4x^2)^(3/2)dx
Let's rewrite the denominator of the integrand,
=intx^2/(-(4x^2-4x+1)+4)^(3/2)dx
=intx^2/((2^2-(2x-1)^2))^(3/2)dx
Now let's use the integral substitution,
Let 2x-1=2sin(theta)
=>2x=1+2sin(theta)
=>x=(1+2sin(theta))/2
dx=1/2(2cos(theta))d theta
dx=cos(theta)d theta
Plug the above in the integral,
=int((1+2sin(theta))/2)^2/(2^2-2^2sin^2(theta))^(3/2)cos(theta)d theta
=int1/4((1+2sin(theta))^2cos(theta))/(2^2(1-sin^2(theta)))^(3/2)d theta
=1/4int((1+2sin(theta))^2cos(theta))/((2^2)^(3/2)(1-sin^2(theta))^(3/2))d theta
=1/4int((1+2sin(theta))^2cos(theta))/(2^3(1-sin^2(theta))^(3/2))d theta
Now use the identity:1-sin^2(x)=cos^2(x)
=1/32int((1+2sin(theta))^2cos(theta))/(cos^2(theta))^(3/2)d theta
=1/32int((1+4sin(theta)+4sin^2(theta))cos(theta))/(cos^3(theta))d theta
=1/32int(1+4sin(theta)+4sin^2(theta))/(cos^2(theta))d theta
=1/32int(1/(cos^2(theta))+(4sin(theta))/(cos^2(theta))+(4sin^2(theta))/(cos^2(theta))d theta
=1/32int(sec^2(theta)+4tan(theta)sec(theta)+4tan^2(theta))d theta
Now use the identity:tan^2(x)=sec^2(x)-1
=1/32int(sec^2(theta)+4tan(theta)sec(theta)+4(sec^2(theta)-1)d theta
=1/32int(5sec^2(theta)+4tan(theta)sec(theta)-4)d theta
Now use the standard integrals,
intsec^2(x)dx=tan(x)+C
intsec(x)tan(x)dx=sec(x)+C
=1/32(5tan(theta)+4sec(theta)-4theta)+C
We have used the integral substitution 2x-1=2sin(theta)
=>sin(theta)=(2x-1)/2
theta=arcsin((2x-1)/2)
Now let's find the tan(theta) and sec(theta) using the right triangle with angle theta and opposite side (2x-1) and hypotenuse as 2,
Use pythagorean identity to find the adjacent side A:
A^2+(2x-1)^2=2^2
A^2+4x^2-4x+1=4
A^2=4-1+4x-4x^2=3+4x-4x^2
A=sqrt(3+4x-4x^2)
tan(theta)=(2x-1)/(sqrt(3+4x-4x^2))
sec(theta)=2/(sqrt(3+4x-4x^2))
Now plug these in the above solution,
=1/32(5*(2x-1)/(sqrt(3+4x-4x^2))+4*2/(sqrt(3+4x-4x^2))-4arcsin((2x-1)/2))+C
=1/32((10x-5+8)/sqrt(3+4x-4x^2)-4arcsin((2x-1)/2))+C
=1/32((10x+3)/sqrt(3+4x-4x^2)-4arcsin((2x-1)/2))+C