Evaluate the integral: I = integral dx/(x(x²+1)²)

Evaluate the integral: $$ I = \int \frac{dx}{x(x^2+1)^2}. $$

Step 1 · Set up partial fractions

$$ \frac{1}{x(x^2+1)^2} = \frac Ax + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}. $$

Multiplying through by $x(x^2+1)^2$: $1=A(x^2+1)^2+(Bx+C)x(x^2+1)+(Dx+E)x$.

Step 2 · Solve for the coefficients

Matching coefficients of $x^4,x^3,x^2,x^1,x^0$ gives $A=1$, $C=0$, $E=0$, $B=-1$, $D=-1$.

Step 3 · Integrate each term

$$ \frac{1}{x(x^2+1)^2} = \frac1x - \frac{x}{x^2+1} - \frac{x}{(x^2+1)^2}. $$

Using $\int x/(x^2+1)\,dx=\tfrac12\ln(x^2+1)$ and $\int x/(x^2+1)^2\,dx=-\tfrac{1}{2(x^2+1)}$:

$$ I = \ln|x| - \frac12\ln\left(x^2+1\right) + \frac{1}{2(x^2+1)} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$, where the denominator vanishes.
• Since $1/[x(x^2+1)^2]$ is an odd function, its decomposition contains only odd-symmetric terms ($1/x$ and pure-$x$-numerator pieces), ruling out $C$ and $E$ from the start.
• The two logarithms combine as $\ln|x|-\tfrac12\ln(x^2+1)=\tfrac12\ln\!\left(\tfrac{x^2}{x^2+1}\right)$.
• The result can be checked by differentiating $\ln|x|-\tfrac12\ln(x^2+1)+\tfrac1{2(x^2+1)}$: this gives $\tfrac1x-\tfrac{x}{x^2+1}-\tfrac{x}{(x^2+1)^2}$, matching the partial-fraction decomposition term by term.