Evaluate the integral: I = integral (x³+1)/(x(x³-8)) dx

Evaluate the integral: $$ I = \int \frac{x^3+1}{x(x^3-8)}\,dx. $$

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 $B=\tfrac38$. Using $(x-2)(x^2+2x+4)=x^3-8$ to simplify the $A$-term, 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 $2x+2=2(x+1)$), the whole term is a direct log-derivative match with no leftover arctangent piece:

$$ \frac34\int \frac{x+1}{x^2+2x+4}\,dx = \frac34\cdot\frac12\ln\left(x^2+2x+4\right) = \frac38\ln\left(x^2+2x+4\right). $$

Step 5 · Combine

$$ I = -\frac18\ln|x| + \frac38\ln|x-2| + \frac38\ln\left(x^2+2x+4\right) + C. $$

Summary Table

Additional Notes

• Correction: a source draft of this problem's solution used incorrect coefficients ($A=\tfrac12$, $B=\tfrac16$, $C=-\tfrac23$, $D=\tfrac56$), which fail to reproduce the original fraction (testing at $x=1$ gives $5/14$ from those coefficients, versus the true value $-2/7$) and led to a spurious arctangent term. That same draft explicitly — and correctly — rejected a "proposed solution" of $\tfrac38\ln|x^3-8|-\tfrac18\ln|x|$ as not matching its own (flawed) derivation; in fact, that rejected proposal is algebraically identical to the correct result above, since $\ln|x^3-8|=\ln|x-2|+\ln(x^2+2x+4)$.
• The domain excludes $x=0$ and $x=2$, the real roots of the denominator ($x^2+2x+4$ has no real roots).
• The result can be checked by differentiating $-\tfrac18\ln|x|+\tfrac38\ln|x-2|+\tfrac38\ln(x^2+2x+4)$: this gives $-\tfrac{1}{8x}+\tfrac{3}{8(x-2)}+\tfrac{3(2x+2)}{8(x^2+2x+4)}$, and evaluating at $x=1$ gives $-\tfrac18-\tfrac38+\tfrac{3\cdot4}{8\cdot7}=-\tfrac12+\tfrac{3}{14}=-\tfrac27$, matching the integrand at $x=1$: $(1+1)/[1\cdot(1-8)]=-\tfrac27$.