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$\int \frac{1}{x^{3}+1} \, dx$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\int \frac{1}{x^{3}+1} \, dx

Simplify using partial fractions: Step $1$ : Simplify the integral using partial fractions. $\newline$ We start by expressing the denominator as a sum of simpler fractions: $\newline$ $x^3 + 1$ can be factored as $(x + 1)(x^2 - x + 1)$ . $\newline$ Now, we set up the partial fractions: $\newline$ $\frac{1}{x^3 + 1} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1}$ .
Solve for A, B, C: Step $2$ : Solve for A, B, and C. $\newline$ Multiply through by $x^3 + 1$ to clear the denominators: $\newline$ $1 = A(x^2 - x + 1) + (Bx + C)(x + 1)$ . $\newline$ Expanding and equating coefficients, we get: $\newline$ $1 = Ax^2 - Ax + A + Bx^2 + Bx + Cx + C$ . $\newline$ Equating coefficients: $\newline$ For $x^2$ : $A + B = 0$ , $\newline$ For $x$ : $-A + B + C = 0$ , $\newline$ For constant: $A + C = 1$ .
Solve system of equations: Step $3$ : Solve the system of equations. $\newline$ From $A + B = 0$ , we get $B = -A$ . $\newline$ Substituting $B = -A$ into $-A + B + C = 0$ gives $-A - A + C = 0$ , so $C = 2A$ . $\newline$ Substituting $C = 2A$ into $A + C = 1$ gives $A + 2A = 1$ , so $A = \frac{1}{3}$ . $\newline$ Then, $B = -A$ $0$ , and $B = -A$ $1$ .
Write integral with coefficients: Step $4$ : Write the integral with the found coefficients. $\newline$ The integral becomes: $\newline$ \int(\(1/(x^ $3$ + $1$ ))\,dx = \int(\frac{ $1$ }{ $3$ })/(x + $1$ )\,dx + \int(\frac{ $-1$ }{ $3$ x} + \frac{ $2$ }{ $3$ })/(x^ $2$ - x + $1$ )\,dx.
Integrate each term: Step $5$ : Integrate each term. $\newline$ For $\int\frac{1}{3}\frac{1}{x + 1}\,dx$ , the integral is $\frac{1}{3}\ln|x + 1|$ . $\newline$ For $\int\frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1}\,dx$ , we need to complete the square in the denominator: $\newline$ $x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$ . $\newline$ The integral becomes more complex and involves a substitution and possibly trigonometric integration.

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