Evaluate the integral: I = integral dx/((x²+1) sqrt(x²-4))

Evaluate the integral: $$ I = \int \frac{dx}{\left(x^2+1\right)\sqrt{x^2-4}}. $$

Step 1 · Secant substitution

For $x>2$, let $x=2\sec\theta$, $0<\theta<\tfrac\pi2$, so $dx=2\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2-4}=2\tan\theta$:

$$ I = \int\frac{2\sec\theta\tan\theta\,d\theta}{\left(4\sec^2\theta+1\right)\cdot2\tan\theta} = \int\frac{\sec\theta\,d\theta}{4\sec^2\theta+1}. $$

Step 2 · Convert to cosines and substitute

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

Let $u=\sin\theta$, $du=\cos\theta\,d\theta$, $\cos^2\theta=1-u^2$:

$$ \int\frac{du}{4+1-u^2} = \int\frac{du}{5-u^2}. $$

Step 3 · Apply the standard formula

With $a=\sqrt5$: $\displaystyle\int\frac{du}{a^2-u^2}=\frac1{2a}\ln\left|\frac{a+u}{a-u}\right|$:

$$ I = \frac1{2\sqrt5}\ln\left|\frac{\sqrt5+\sin\theta}{\sqrt5-\sin\theta}\right| + C. $$

Step 4 · Convert back to $x$

Since $x=2\sec\theta$, $\cos\theta=2/x$ and $\sin\theta=\sqrt{x^2-4}/x$ for $x>2$:

$$ I = \frac1{2\sqrt5}\ln\left|\frac{x\sqrt5+\sqrt{x^2-4}}{x\sqrt5-\sqrt{x^2-4}}\right| + C. $$

Summary Table

Additional Notes

• The domain requires $|x|>2$; the derivation shown covers $x>2$ (the $x<-2$ branch follows by symmetry).
• The result was verified numerically at $x=3$: the chain-rule derivative evaluates to $\sqrt5/50\approx0.04472$, matching the integrand $1/(10\sqrt5)\approx0.04472$ there.