Evaluate the integral: I = integral (x³+1)/(x(x³-8)) dx
Evaluate the integral: $$ I = \int \frac{x^3+1}{x(x^3-8)}\,dx. $$
Explanation of Method
This is the same integral as Problem 262. Factoring the cubic denominator into a linear factor and an irreducible quadratic, then decomposing into partial fractions, reduces the integral to logarithms — with no arctangent term needed, since the quadratic-factor numerator turns out to be an exact multiple of the denominator's derivative.
- Factor $x^3-8=(x-2)(x^2+2x+4)$.
- Decompose into partial fractions.
- Solve for the coefficients.
- Integrate each term.
Step 1 · Factor and set up partial fractions
$$ \frac{x^3+1}{x(x-2)(x^2+2x+4)} = \frac Ax + \frac{B}{x-2} + \frac{Cx+D}{x^2+2x+4}. $$
Multiplying through: $x^3+1=A(x-2)(x^2+2x+4)+Bx(x^2+2x+4)+(Cx+D)x(x-2)$.
Step 2 · Solve for the coefficients
Setting $x=0$: $1=A(-2)(4)=-8A$, so $A=-\tfrac18$. Setting $x=2$: $9=B(2)(12)=24B$, so $B=\tfrac38$. Matching the remaining coefficients gives $C=\tfrac34$ and $D=\tfrac34$.
Step 3 · Integrate the linear-factor terms
$$ -\frac18\int \frac{dx}{x} + \frac38\int \frac{dx}{x-2} = -\frac18\ln|x| + \frac38\ln|x-2|. $$
Step 4 · Integrate the quadratic-factor term
Since $Cx+D=\tfrac34(x+1)$ is exactly proportional to $\tfrac12$ the derivative of $x^2+2x+4$ (which is $2x+2=2(x+1)$), this term is a direct log-derivative match with no leftover arctangent piece:
$$ \frac34\int \frac{x+1}{x^2+2x+4}\,dx = \frac38\ln\left(x^2+2x+4\right). $$
Step 5 · Combine
$$ I = -\frac18\ln|x| + \frac38\ln|x-2| + \frac38\ln\left(x^2+2x+4\right) + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ | $-1/8$ |
| $B$ | $3/8$ |
| $C$ | $3/4$ |
| $D$ | $3/4$ |
Additional Notes
- Correction: the source draft (repeated identically for Problems 262 and 263) used incorrect coefficients ($A=\tfrac12,B=\tfrac16,C=-\tfrac23,D=\tfrac56$), which fail to reproduce the original fraction and led to a spurious arctangent term. That same draft explicitly — and correctly — rejected a "proposed solution" of $\tfrac38\ln|x^3-8|-\tfrac18\ln|x|$; that rejected proposal is in fact algebraically identical to the correct result above.
- The domain excludes $x=0$ and $x=2$, the real roots of the denominator.
- The result can be checked by differentiating $-\tfrac18\ln|x|+\tfrac38\ln|x-2|+\tfrac38\ln(x^2+2x+4)$, which evaluates at $x=1$ to $-\tfrac27$, matching the integrand $(1+1)/[1(1-8)]=-\tfrac27$.
Result
The solution to the integral is: