Question

Evaluate the integral: \[ \int \frac{x^3 - 1}{x^3 + x} dx \]

• A. \( x + \log|x| + \frac{1}{2} \log(x^2 + 1) + \sin^{-1}(x) + c \)
• B. \( x - \log|x| + \frac{1}{2} \log(x^2 + 1) - \sin^{-1}(x) + c \)
• C. \( x + \log|x| - \frac{1}{2} \log(x^2 + 1) + \tan^{-1}(x) + c \)
• D. \( x - \log|x| + \frac{1}{2} \log(x^2 + 1) - \tan^{-1}(x) + c \)

Hint:

Remember to break complex rational expressions into simpler parts for easier integration using partial fraction decomposition.

Answer 1

The Correct Option is D

The integrand is rewritten as:\[\frac{x^3 - 1}{x^3 + x} = \frac{x^3 + x - x - 1}{x^3 + x} = 1 - \frac{x + 1}{x(x^2 + 1)}\]The remaining fraction is decomposed using partial fractions:\[\frac{x + 1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}\]\[x + 1 = A(x^2 + 1) + (Bx + C)x\]Setting \(x = 0\) yields:\[1 = A\]Comparing coefficients of \(x^2\) gives:\[0 = A + B \implies B = -1\]Comparing coefficients of \(x\) yields:\[1 = C\]Thus,\[\frac{x + 1}{x(x^2 + 1)} = \frac{1}{x} - \frac{x}{x^2 + 1} + \frac{1}{x^2 + 1}\]The integral is computed as follows:\begin{align*}\int \frac{x^3 - 1}{x^3 + x} dx &= \int \left(1 - \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{1}{x^2 + 1}\right) dx
&= x - \ln|x| + \frac{1}{2} \ln(x^2 + 1) - \tan^-^1(x) + c\end{align*}