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

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

Step 1 · Set up partial fractions

Since the integrand is an even function, only even-symmetric terms can appear:

$$ \frac{1}{x^2(1+x^2)^2} = \frac{B}{x^2} + \frac{D}{1+x^2} + \frac{F}{(1+x^2)^2}. $$

Multiplying through by $x^2(1+x^2)^2$: $1=B(1+x^2)^2+Dx^2(1+x^2)+Fx^2$.

Step 2 · Solve for the coefficients

Expanding and matching coefficients of $x^4$, $x^2$, and the constant term:

$$ B+D=0, \qquad 2B+D+F=0, \qquad B=1. $$

From $B=1$: $D=-1$. Substituting into the middle equation: $2-1+F=0$, so $F=-1$.

Step 3 · Integrate each term

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

Using $\int dx/x^2=-1/x$, $\int dx/(1+x^2)=\arctan x$, and the standard reduction result $\int dx/(1+x^2)^2=\tfrac{x}{2(1+x^2)}+\tfrac12\arctan x$:

$$ I = -\frac1x - \arctan x - \left[\frac{x}{2(1+x^2)}+\frac12\arctan x\right] + C. $$

Step 4 · Combine

$$ I = -\frac1x - \frac{x}{2(1+x^2)} - \frac32\arctan x + C. $$

Summary Table

Additional Notes

• Correction: the source claimed $D=-\tfrac12$ and $F=-\tfrac12$, but plugging these into the decomposition and testing at $x=1$ gives $1-\tfrac14-\tfrac18=\tfrac58$, not the required $\tfrac14=1/[1^2(1+1^2)^2]$. Re-solving the coefficient-matching system directly (from $B+D=0$, $2B+D+F=0$, $B=1$) gives $D=-1$ and $F=-1$ instead, which checks out at $x=1$: $1-\tfrac12-\tfrac14=\tfrac14$. This error carried through to make the source's final answer exactly half the correct size on the two non-power-rule terms.
• The domain excludes $x=0$, where the denominator vanishes.
• The result can be checked by differentiating $-\tfrac1x-\tfrac{x}{2(1+x^2)}-\tfrac32\arctan x$: combining all three derivative terms over $2x^2(1+x^2)^2$ gives numerator $2(1+x^2)^2+x^2(x^2-1)-3x^2(1+x^2)$, which simplifies to the constant $2$, reducing the whole expression to exactly $1/[x^2(1+x^2)^2]$.