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

We can rewrite the integrand as: \[ \frac{x^3 - 1}{x^3 + x} = \frac{x^3 + x - x - 1}{x^3 + x} = 1 - \frac{x + 1}{x(x^2 + 1)} \] Now, we use partial fraction decomposition on the remaining fraction: \[ \frac{x + 1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \] \[ x + 1 = A(x^2 + 1) + (Bx + C)x \] Let \(x = 0\): \[ 1 = A \] Comparing coefficients of \(x^2\): \[ 0 = A + B \implies B = -1 \] Comparing coefficients of \(x\): \[ 1 = C \] So, \[ \frac{x + 1}{x(x^2 + 1)} = \frac{1}{x} - \frac{x}{x^2 + 1} + \frac{1}{x^2 + 1} \] Now, we integrate: \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*}

Answer 2

Step 1: Start with the given integral
We are tasked with evaluating the following integral: \[ I = \int \frac{x^3 - 1}{x^3 + x} \, dx. \]
Step 2: Simplify the integrand
First, factor the denominator: \[ x^3 + x = x(x^2 + 1). \] Thus, the integral becomes: \[ I = \int \frac{x^3 - 1}{x(x^2 + 1)} \, dx. \]
Step 3: Decompose the rational function
We can decompose the integrand into simpler terms. We split the numerator \( x^3 - 1 \) as follows: \[ x^3 - 1 = (x - 1)(x^2 + x + 1). \] Therefore, the integral becomes: \[ I = \int \frac{(x - 1)(x^2 + x + 1)}{x(x^2 + 1)} \, dx. \] Now split the rational function into two parts: \[ I = \int \left( \frac{x^2 + x + 1}{x^2 + 1} \, dx - \frac{1}{x(x^2 + 1)} \, dx \right). \]
Step 4: Solve the two integrals
1. For the first integral, simplify the expression: \[ \int \frac{x^2 + x + 1}{x^2 + 1} \, dx = \int 1 \, dx + \int \frac{x}{x^2 + 1} \, dx = x + \frac{1}{2} \log(x^2 + 1). \] 2. For the second integral, use substitution: \[ \int \frac{1}{x(x^2 + 1)} \, dx. \] This is a standard integral and results in: \[ - \log|x| + \tan^{-1}(x). \]
Step 5: Combine the results
Now combine the results of both integrals: \[ I = x - \log|x| + \frac{1}{2} \log(x^2 + 1) - \tan^{-1}(x) + c. \] Thus, the value of the integral is: \[ \boxed{x - \log|x| + \frac{1}{2} \log(x^2 + 1) - \tan^{-1}(x) + c.} \]