Evaluate the integral: I = integral dx/(x^4+3x²)

Evaluate the integral: $$ I = \int \frac{dx}{x^4+3x^2}. $$

Step 1 · Factor and set up partial fractions

$$ x^4+3x^2 = x^2(x^2+3), \qquad \frac{1}{x^2(x^2+3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+3}. $$

Multiplying through by $x^2(x^2+3)$:

$$ 1 = Ax(x^2+3) + B(x^2+3) + (Cx+D)x^2 = (A+C)x^3 + (B+D)x^2 + 3Ax + 3B. $$

Step 2 · Match coefficients

Comparing to $0x^3+0x^2+0x+1$: $A+C=0$, $B+D=0$, $3A=0$, and $3B=1$. Solving: $A=0$, $B=\tfrac13$, $C=0$, $D=-\tfrac13$.

Step 3 · Integrate each term

$$ \frac{1}{x^2(x^2+3)} = \frac{1/3}{x^2} - \frac{1/3}{x^2+3}, \qquad I = \frac13\int \frac{dx}{x^2} - \frac13\int \frac{dx}{x^2+3}. $$

Using $\int dx/x^2=-1/x$ and the standard arctangent formula $\int dx/(x^2+a^2)=\tfrac1a\arctan(x/a)$ with $a=\sqrt3$:

$$ I = \frac13\left(-\frac1x\right) - \frac13\cdot\frac{1}{\sqrt3}\arctan\left(\frac{x}{\sqrt3}\right) + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$, the repeated root of the factor $x^2$; $x^2+3$ contributes no real roots.
• Since $A=C=0$, the decomposition has no $1/x$ or $x/(x^2+3)$ terms — only the two even-symmetric pieces survive, which makes sense since the original integrand $1/(x^4+3x^2)$ is an even function.
• The result can be checked by differentiating $-\tfrac{1}{3x}-\tfrac1{3\sqrt3}\arctan(x/\sqrt3)$: this gives $\tfrac{1}{3x^2}-\tfrac13\cdot\tfrac{1}{3}\cdot\tfrac{3}{3+x^2}=\tfrac{1}{3x^2}-\tfrac{1}{3(x^2+3)}$, and combining over $3x^2(x^2+3)$ gives numerator $(x^2+3)-x^2=3$, so the whole expression reduces to $\tfrac{3}{3x^2(x^2+3)}=\tfrac{1}{x^2(x^2+3)}$, matching the integrand.