Hello, please help me with my intergral

Question

So the equation goes something like this, it is a fraction

the integral of (x^4 + X^2 + 4) over (x^6 +4x^4 + 4x^2) then dx on the middle of the fraction

Please show me how to solve it and provide me some explanation so I cannot understand it and further study it. Thank you so much. Have a great day!

Answer 1

I = ∫(x⁴ + X² + 4)/(x⁶ +4x⁴ + 4x²∫)dx = ∫(x⁴ + x²+ 4)/[x²(x² + 2)²]dx

You have to decompouse this rational function: (x⁴ + x²+ 4)/[x²(x² + 2)²] = a/x² + b/x + (cx + d)/(x² + 2)² +

(ex + f)/(x² + 2); x⁴ + x²+ 4 = a(x² + 2)² + bx(x² + 2)² + (cx + d)·x² + (ex + f)x²(x² + 2)

b + e = 0, e = 0

a + f = 1, f = 0

4b + c = 0, c = 0

4a + d + 2f = 1, d = - 3

4b = 0, b = 0

4a = 4, a = 1

I = ∫(1/x² - 3/(x² + 2)²)dx = - 1/x - 3∫dx/(x² + 2)². In last integral let x = √2tanu; dx = √2sec²udu

-3∫dx/(x² + 2)² = -3∫√2sec²udu/4sec⁴u = - 3√2/4∫cos²udu = -3√2/8(u + 1/2sin2u) = - 3√2/8(tan^-1x/√2 + x√2/(x² + 2). So our integral I = -1/x -3√2/8(tan^-1x/√2 + x√2/(x²+ 2)) + C