Question

Evaluating \( \int \frac{dx}{x^3 + 1} \):

Hint:

When dealing with integrals involving sums of cubes, factor the denominator and use partial fractions to decompose the expression.

Answer 1

Step 1: Factor the denominator.
The denominator \( x^3 + 1 \) is a sum of cubes, so we can factor it as: \[ x^3 + 1 = (x + 1)(x^2 - x + 1) \] Thus, the integral becomes: \[ \int \frac{dx}{(x + 1)(x^2 - x + 1)} \] Step 2: Use partial fractions.
To decompose the fraction, we use partial fractions. We express \( \frac{1}{(x + 1)(x^2 - x + 1)} \) as: \[ \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} \] Multiply both sides by \( (x + 1)(x^2 - x + 1) \): \[ 1 = A(x^2 - x + 1) + (Bx + C)(x + 1) \] Expand both sides: \[ 1 = A(x^2 - x + 1) + (Bx^2 + Bx + Cx + C) \] Simplify: \[ 1 = (A + B)x^2 + (B + C)x + (A + C) \] Now, equate the coefficients of like powers of \( x \):
- For \( x^2 \), \( A + B = 0 \)
- For \( x \), \( B + C = 0 \)
- For the constant term, \( A + C = 1 \)
Solving these equations, we find: \[ A = 1, \quad B = -1, \quad C = 1 \] Thus, the partial fraction decomposition is: \[ \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{1}{x + 1} - \frac{x - 1}{x^2 - x + 1} \] Step 3: Integrating the terms.
Now, integrate each term: \[ \int \frac{dx}{x + 1} = \ln |x + 1| \] \[ \int \frac{x - 1}{x^2 - x + 1} \, dx \] This can be integrated using a standard substitution. Let \( u = x^2 - x + 1 \), so \( du = (2x - 1) dx \). This simplifies the integral to: \[ \int \frac{du}{u} = \ln |u| = \ln |x^2 - x + 1| \] Step 4: Final Answer.
Thus, the integral becomes: \[ \int \frac{dx}{x^3 + 1} = \ln |x + 1| - \ln |x^2 - x + 1| + C \]