Question

If $$ \int \frac{x^4 + 1}{x^2 + 1} dx = Ax^3 + Bx^2 + Cx + D \tan^{-1} x + E, $$ then find $ A + B + C + D $.

• A. \( \frac{3}{2} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{2}{3} \)

Hint:

Use polynomial division and standard integrals.

Answer 1

The Correct Option is B

Divide numerator by denominator: \[ \frac{x^4 + 1}{x^2 + 1} = x^2 - 1 + \frac{2}{x^2 + 1} \] Integral: \[ \int (x^2 - 1) dx + \int \frac{2}{x^2 + 1} dx = \frac{x^3}{3} - x + 2 \tan^{-1} x + C \] Compare with given: \[ Ax^3 + Bx^2 + Cx + D \tan^{-1} x + E \] So: \[ A = \frac{1}{3}, B = 0, C = -1, D = 2 \] Sum: \[ A + B + C + D = \frac{1}{3} + 0 - 1 + 2 = \frac{4}{3} \]