Evaluate the integral: I = integral dx/((a²+x²)²)

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

Step 1 · Trigonometric substitution

Let $x=a\tan\theta$, so $dx=a\sec^2\theta\,d\theta$ and $a^2+x^2=a^2\sec^2\theta$:

$$ I = \int \frac{a\sec^2\theta\,d\theta}{a^4\sec^4\theta} = \frac{1}{a^3}\int \cos^2\theta\,d\theta. $$

Step 2 · Power-reduction identity

Using $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$:

$$ I = \frac{1}{2a^3}\left(\theta + \frac{\sin2\theta}{2}\right) + C. $$

Step 3 · Substitute back

With $\theta=\arctan(x/a)$ and $\sin(2\arctan u)=\dfrac{2u}{1+u^2}$, $u=x/a$:

$$ \sin2\theta = \frac{2(x/a)}{1+(x/a)^2} = \frac{2ax}{a^2+x^2}. $$

Substituting this in:

$$ I = \frac{1}{2a^3}\left(\arctan\frac xa + \frac12\cdot\frac{2ax}{a^2+x^2}\right) + C = \frac{1}{2a^3}\arctan\frac xa + \frac{x}{2a^2(a^2+x^2)} + C. $$

Summary Table

Additional Notes

• Correction: the source's final simplification wrote the back-substituted term as $x/(2(a^2+x^2))$, dropping the factor of $a$ that appears in $\sin(2\arctan u)=2u/(1+u^2)=2ax/(a^2+x^2)$ once $u=x/a$ is converted to a single fraction in $x$ and $a$. The correct term is $x/(2a^2(a^2+x^2))$, matching the well-known reduction formula for $\int dx/(a^2+x^2)^2$.
• The integrand is defined for all real $x$ (assuming $a\neq0$).
• The result can be checked by differentiating $\tfrac{x}{2a^2(a^2+x^2)}+\tfrac{1}{2a^3}\arctan(x/a)$: the quotient-rule derivative of the first term is $\tfrac{a^2-x^2}{2a^2(a^2+x^2)^2}$, and the derivative of the second term is $\tfrac{1}{2a^2(a^2+x^2)}=\tfrac{a^2+x^2}{2a^2(a^2+x^2)^2}$; summing gives $\tfrac{2a^2}{2a^2(a^2+x^2)^2}=\tfrac{1}{(a^2+x^2)^2}$, matching the integrand.