Question

If \( \int \frac{x^8+4}{x^4-2x^2+2} dx = Ax^5+Bx^3+Cx+K \), then \(5A+3B+C = \)

• A. 7
• B. 5
• C. 3
• D. 1

Hint:

Look for algebraic simplifications before integrating rational functions.
Sophie Germain Identity: \(a^4+4b^4 = (a^2+2b^2+2ab)(a^2+2b^2-2ab)\).
If the integrand simplifies to a polynomial, integration is straightforward.
Compare coefficients to find A, B, C.

Answer 1

The Correct Option is B

The integrand is \( f(x) = \frac{x^8+4}{x^4-2x^2+2} \). This is an improper rational function since degree of numerator (8)>degree of denominator (4). We need to perform polynomial long division. Notice that \(x^4-2x^2+2\) looks like it might be part of \( (x^2-1)^2+1 = x^4-2x^2+1+1 \). Also note \(x^8+4 = (x^4)^2+2^2\). We can use Sophie Germain identity: \(a^4+4b^4 = (a^2+2b^2+2ab)(a^2+2b^2-2ab)\). Let \(a=x^2, 4b^4=4 \implies b^4=1 \implies b=1\). So \(x^8+4 = ( (x^2)^2 + 2(1)^2 + 2(x^2)(1) ) ( (x^2)^2 + 2(1)^2 - 2(x^2)(1) ) \) \(x^8+4 = (x^4+2x^2+2)(x^4-2x^2+2)\). Therefore, the integrand simplifies: \[ \frac{x^8+4}{x^4-2x^2+2} = \frac{(x^4+2x^2+2)(x^4-2x^2+2)}{x^4-2x^2+2} = x^4+2x^2+2 \] (Assuming \(x^4-2x^2+2 \neq 0\). The roots of \(y^2-2y+2=0\) where \(y=x^2\) are \(y = \frac{2 \pm \sqrt{4-8}}{2} = 1 \pm i\). So \(x^2=1\pm i\), which means denominator is never zero for real x.) Now we integrate this simplified polynomial: \( \int (x^4+2x^2+2) dx = \frac{x^5}{5} + \frac{2x^3}{3} + 2x + K' \) (using K' for integration constant). This is given to be equal to \(Ax^5+Bx^3+Cx+K\). Comparing coefficients: \(A = 1/5\) \(B = 2/3\) \(C = 2\) We need to find \(5A+3B+C\): \(5A+3B+C = 5(1/5) + 3(2/3) + 2\) \( = 1 + 2 + 2 = 5 \). This matches option (b). \[ \boxed{5} \]