`int x^2 /(3 + 4x - 4x^2)^(3/2) dx` Evaluate the integral

`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`