Evaluate the following integrals SolutionI integral of x16x

Solution

I = integral of x[(16x^2-9) + x/(9-16x^2)]dx
I = integral of [x(16x^2-9) + x^2/(9-16x^2)]dx
lets separate both integrals as
I = I1 + I2
now solve them individually
I1 = integral of [x(16x^2-9)]dx
let 16x^2-9 = t
32xdx = dt
xdx= dt/32
I1 = 1/32 t dt
integrating
I1 = 1/32 * 2/3 t^3/2
I1 = 1/48* (16x^2-9)^3/2
apply limit from 3/4 to 32/4
I1 = 1/48 [(16*18/16 - 9)^3/2 - (16*9/16 - 9)^3/2]
I1 = 1/48[27]
I1 = 9/16.......(1)
I2 = x^2/(9-16x^2)dx
I2 = 1/16 * 16x^2/(9-16x^2)
I2 = 1/16 * (16x^2-9+9)/(9-16x^2)
I2 = 1/16 * [-(9-16x^2) + 9/(9-16x^2)]dx
I2 = 1/16*4[-(9/16-x^2) + 9/(9/16-x^2)]dx
I2 = 1/4[-(9/16-x^2) + 9/(9/16-x^2)]dx
integrating wrt x
I2 = 1/4 [-x/2(9/16-x^2) - 9/32sin^-1(4x/3) + 9sin^-1(4x/3)]
apply limits
I2 = 1/4[-9/32sin^-1(2) + 9sin^-1(2) + 9/32/2 - 9/2]
I2 = 1/4[279/32sin^-1(2) - 279/32*/2]
I = I1 + I2
I = 9/16 + 1/4[279/32sin^-1(2) - 279/32*/2]