Evaluate the integral: I = integral dx/((x²+1) sqrt(x²-1))
Evaluate the integral: $$ I = \int \frac{dx}{(x^2+1)\sqrt{x^2-1}}. $$
Explanation of Method
The trig substitution $x=\sec\theta$ collapses $\sqrt{x^2-1}=\tan\theta$, reducing the integral to $\int\cos\theta\,d\theta/(1+\cos^2\theta)$, which a $u=\sin\theta$ substitution turns into a partial-fractions integral.
- Substitute $x=\sec\theta$.
- Simplify to $\int\cos\theta\,d\theta/(1+\cos^2\theta)$.
- Substitute $u=\sin\theta$ and apply partial fractions.
- Back-substitute.
Step 1 · Substitute
Let $x=\sec\theta$, so $dx=\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2-1}=\tan\theta$:
$$ I = \int\frac{\sec\theta\tan\theta\,d\theta}{(\sec^2\theta+1)\tan\theta} = \int\frac{\sec\theta\,d\theta}{\sec^2\theta+1}. $$
Step 2 · Simplify
Multiplying numerator and denominator by $\cos^2\theta$:
$$ I = \int\frac{\cos\theta\,d\theta}{1+\cos^2\theta}. $$
Step 3 · Substitute and integrate
Let $u=\sin\theta$, so $\cos^2\theta=1-u^2$, and:
$$ I = \int\frac{du}{2-u^2} = \frac{\sqrt2}4\ln\left|\frac{\sqrt2+u}{\sqrt2-u}\right| + C. $$
Step 4 · Back-substitute
With $x=\sec\theta$, $\sin\theta=\sqrt{x^2-1}/x$; multiplying inside the log by $x$:
$$ I = \frac{\sqrt2}4\ln\left|\frac{\sqrt2x+\sqrt{x^2-1}}{\sqrt2x-\sqrt{x^2-1}}\right| + C. $$
Summary Table
| Substitution | Effect |
|---|---|
| $x=\sec\theta$ | reduces to $\int\cos\theta\,d\theta/(1+\cos^2\theta)$ |
| $u=\sin\theta$ | reduces to $\int du/(2-u^2)$ |
Additional Notes
- The domain is $x>1$ (or $x<-1$ by symmetry).
- The result was verified by differentiation at $x=2$ and $x=3$: the derivative matches the integrand $1/[(x^2+1)\sqrt{x^2-1}]$ exactly at both points.
Result
The solution to the integral is: