Can you answer this Calculus question

∫[(3x^2-2x+4)/(x^3+2x)]

Decompose the original integrand into partial fractions:
(3x² - 2x + 4) / (x³ + 2x) = x(x² + 2)
(3x² - 2x + 4) / (x³ + 2x) = A / x + (Bx + C) / (x² + 2)
3x² - 2x + 4 = A(x² + 2) + x(Bx + C)
3x² - 2x + 4 = Ax² + 2A + Bx² + Cx
3x² - 2x + 4 = (A + B)x² + Cx + 2A
2A = 4
A = 2
A + B = 3
B = 3 - A
B = 3 - 2
B = 1
C = -2
(3x² - 2x + 4) / (x³ + 2x) = 2 / x + (x - 2) / (x² + 2)
(3x² - 2x + 4) / (x³ + 2x) = 2 / x + x / (x² + 2) - 2 / (x² + 2)

Integrate the new integrand using the partial fractions:
∫ (3x² - 2x + 4) / (x³ + 2x) dx = ∫ [2 / x + x / (x² + 2) - 2 / (x² + 2)] dx
∫ (3x² - 2x + 4) / (x³ + 2x) dx = 2 ∫ 1 / x dx + ∫ 2x / (x² + 2) dx / 2 - 2 ∫ 1 / (x² + 2) dx
∫ (3x² - 2x + 4) / (x³ + 2x) dx = 2ln|x| + ln(x² + 2) / 2 - √2tanˉ¹(x / √2) + C
∫ (3x² - 2x + 4) / (x³ + 2x) dx = lnx² + ln√(x² + 2) - √2tanˉ¹(x / √2) + C
∫ (3x² - 2x + 4) / (x³ + 2x) dx = ln[x²√(x² + 2)] - √2tanˉ¹(x / √2) + C

∫[(3x^2-2x+4)/(x^3+2x) dx = ∫[(3x^2-2x+4)/x(x^2+2) dx

Decompose into partial fractions.

3x^2-2x+4/x(x^2+2) = A/x+ (Bx+C)/(x^2+2)

3x^2-2x+4 = A(x^2+2) + (Bx+C)x

compare the coefficients of x^2 on both sides
3 = A + B -----(1)
compare the coefficients of x
-2 = C ----(2)
compare the constant terms
4 = 2A
A = 2
Therefore B=1

A/x+ (Bx+C)/(x^2+2) = 2/x + (x-2) /(x^2+2)

∫[(3x^2-2x+4)/(x^3+2x) dx = ∫[2/x + (x-2) /(x^2+2)] dx

= ∫ 2 dx /x + ∫ x dx /(x^2+2) - 2 ∫ dx/(x^2+2) -----------(A)

∫ 2 dx /x = 2 ln x ------(1)

∫ x dx /(x^2+2) = (1/2) ∫ 2x dx /(x^2+2)
= (1/2) ln (x^2+2) --------(2)

2 ∫ dx/(x^2+2) = -2 (1/√2) tan^-1 (x/√2) --- (3)

(A) = (1)+(2) + (3)
= 2 ln x + (1/2) ln (x^2+2) - √2 tan^-1 (x/√2) + C