Question

$\displaystyle \int \frac{x^4 - 1}{x + 1} \, dx$ is equal to:

• A. $\frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + x + C$
• B. $\frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + 2x + C$
• C. $\frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} - x + C$
• D. $\frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} - 2x + C$
• E. $-\frac{x^4}{4} - \frac{x^3}{3} - \frac{x^2}{2} - x + C$

Hint:

Always simplify rational functions using division before integrating.

Answer 1

The Correct Option is C

Concept:
• Use polynomial division before integrating

Step 1: Divide $x^4 - 1$ by $x+1$
\[ \frac{x^4 - 1}{x+1} = x^3 - x^2 + x - 1 \]

Step 2: Integrate term-wise
\[ \int (x^3 - x^2 + x - 1)\,dx \] \[ = \frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} - x + C \] Final Conclusion:
Option (C)