We have to solve the equation: Int[(sin x)^2 dx]+Int[(cos x)^2 dx] = (x^3-4x)/(x-2)

Now Int [ a dx ] + Int [ b dx] = Int [ (a + b) dx]

Int[(sin x)^2 dx]+Int[(cos x)^2 dx] = (x^3 - 4x)/(x-2)

=> Int [ (sin x)^2 + (cos x)^2 dx] = (x^3 - 4x)/(x-2)

Use (sinx)^2 + (cos x)^2 = 1

=> Int [ 1 dx] = (x^3 - 4x)/(x-2)

=> x = (x^3 - 4x)/(x-2)

=> x = x(x^2 - 4)/(x-2)

=> x - 2 = x^2 - 4

=> x^2 - x - 2 = 0

=> x^2 - 2x + x - 2 = 0

=> x(x - 2) + 1(x - 2) = 0

=> (x + 1)(x - 2) = 0

=> x = -1 and x = 2

But x = 2 makes the left hand side undefined, so we do not consider it.

**The required solution of the equation is x = -1.**

We'll re-write the etrms from the left side of the equation, solving the integrals.

We'll use the half angle identities:

[cos (a/2)]^2 = (1 + cos a)/2

[sin (a/2)]^2 = (1 - cos a)/2

Int (sin x)^2 dx = (1/2)Int dx - (1/2)Int cos 2x dx = x/2 - sin 2x/4 +c

Int (cos x)^2 dx = x/2 + sin 2x/4 + c

Now, we'll solve the equation substituting the integrals by their results:

x/2 - sin 2x/4 + x/2 + sin 2x/4 = (x^3-4x)/(x-2)

We'll eliminate like terms and we'll factorize by x to the right side:

2x/2 = x(x^2-4)/(x-2)

We'll re-write the difference of squares from numerator: x^2 - 4 = (x-2)(x+2)

2x/2 = x(x-2)(x+2)/(x-2)

We'll simplify and we'll get:

x = x(x+2)

We'll divide by x:

x+2=1

We'll subtract 2 both sides:

x=-1

**The solution of the equation is x = -1.**