Evaluate the integral: I = integral (x²)/((x²+1)³) dx

Evaluate the integral: $$ I = \int \frac{x^2}{\left(x^2+1\right)^3}\,dx. $$

Step 1 · Rewrite and split

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

Step 2 · Apply the reduction formula

Using $\displaystyle\int\frac{dx}{(x^2+1)^2}=\frac{x}{2(x^2+1)}+\frac12\arctan x$ and, applying the reduction formula once more,

$$ \int\frac{dx}{(x^2+1)^3} = \frac{x}{4(x^2+1)^2} + \frac34\int\frac{dx}{(x^2+1)^2} = \frac{x}{4(x^2+1)^2} + \frac{3x}{8(x^2+1)} + \frac38\arctan x. $$

Step 3 · Combine

$$ I = \left[\frac{x}{2(x^2+1)}+\frac12\arctan x\right] - \left[\frac{x}{4(x^2+1)^2} + \frac{3x}{8(x^2+1)} + \frac38\arctan x\right]. $$

Collecting the $\arctan x$ terms: $\tfrac12-\tfrac38=\tfrac18$. Collecting the $x/(x^2+1)$ terms: $\tfrac{x}{2(x^2+1)}-\tfrac{3x}{8(x^2+1)}=\tfrac{x}{8(x^2+1)}$. Combining this with $-\tfrac{x}{4(x^2+1)^2}$ over the common denominator $8(x^2+1)^2$:

$$ \frac{x}{8(x^2+1)} - \frac{x}{4(x^2+1)^2} = \frac{x(x^2+1)-2x}{8(x^2+1)^2} = \frac{x^3-x}{8(x^2+1)^2} = \frac{x(x^2-1)}{8(x^2+1)^2}. $$

Summary Table

Additional Notes

• The source's final answer and derivation were correct; the version above presents the same reduction-formula approach with the algebra shown explicitly.
• The domain is all real $x$, since $x^2+1>0$ everywhere.
• The result can be checked by differentiating $\tfrac{x(x^2-1)}{8(x^2+1)^2}+\tfrac18\arctan x$ via the quotient rule: this gives $\tfrac{-x^4+6x^2-1}{8(x^2+1)^3}+\tfrac1{8(x^2+1)}$, and combining over $(x^2+1)^3$ gives numerator $-x^4+6x^2-1+(x^2+1)^2=8x^2$, so the derivative is $\tfrac{8x^2}{8(x^2+1)^3}=\tfrac{x^2}{(x^2+1)^3}$, matching the integrand exactly.