Evaluate the integral: I = integral dx/(x(x³+4))

Evaluate the integral: $$ I = \int \frac{dx}{x(x^3+4)}. $$

Step 1 · Split the integrand

Since $(x^3+4)-x^3=4$, dividing both sides by $4x(x^3+4)$ gives the identity:

$$ \frac{1}{x(x^3+4)} = \frac14\left(\frac1x - \frac{x^2}{x^3+4}\right). $$

Step 2 · Integrate each piece

$$ I = \frac14\int \frac{dx}{x} - \frac14\int \frac{x^2}{x^3+4}\,dx. $$

The first piece is $\tfrac14\ln|x|$. For the second, let $w=x^3+4$, $dw=3x^2\,dx$:

$$ \frac14\int \frac{x^2}{x^3+4}\,dx = \frac{1}{12}\int \frac{dw}{w} = \frac{1}{12}\ln|x^3+4|. $$

Step 3 · Combine

$$ I = \frac14\ln|x| - \frac{1}{12}\ln|x^3+4| + C = \frac{1}{12}\ln\left|\frac{x^3}{x^3+4}\right| + C. $$

Summary Table

Additional Notes

• Correction: the source claimed $x^3+4=x^3+2^3$, but $2^3=8$, not $4$ — this "sum of cubes" framing was incorrect. Fortunately the actual splitting identity used ($1/[x(x^n+a)]=\tfrac1a[1/x-x^{n-1}/(x^n+a)]$) holds for any nonzero constant $a$, not just perfect cubes, so the rest of the source's computation (and its final answer) was unaffected by this framing error; the identity was independently re-derived above without reference to any cube-root structure.
• The domain excludes $x=0$ and $x=-\sqrt[3]4$, where the denominator vanishes.
• The result can be checked by differentiating $\tfrac1{12}\ln\left|\tfrac{x^3}{x^3+4}\right|=\tfrac1{12}(3\ln|x|-\ln|x^3+4|)=\tfrac14\ln|x|-\tfrac1{12}\ln|x^3+4|$: this gives $\tfrac{1}{4x}-\tfrac{3x^2}{12(x^3+4)}=\tfrac{1}{4x}-\tfrac{x^2}{4(x^3+4)}=\tfrac{(x^3+4)-x^3}{4x(x^3+4)}=\tfrac{4}{4x(x^3+4)}=\tfrac{1}{x(x^3+4)}$, matching the integrand.