Q.Evaluate $\int_{\log\sqrt{2}}^{\log\sqrt{3}} \frac{1}{(e^x + e^{-x})(e^x - e^{-x})}\, dx$.

The problem asks for $\displaystyle\int \frac{1}{x}\sqrt{\frac{x + a}{x - a}}\, dx$. At first glance, the square root of a rational function suggests a substitution that will remove the radical. The expression inside the root, $\frac{x + a}{x - a}$, is reminiscent of trigonometric identities — specifically, the secant function.

Recall that $\sec^2\theta - 1 = \tan^2\theta$. If we set $x = a \sec\theta$, then $x^2 - a^2 = a^2 \tan^2\theta$, and the square root will simplify nicely. This is a classic technique for integrals involving $\sqrt{x^2 - a^2}$ or similar forms.

Let's walk through it step by step.

1. Choose the substitution.

   Let $x = a \sec\theta$, where $a > 0$ (we can handle sign later). Then $dx = a \sec\theta \tan\theta \, d\theta$.

   The domain: $x > a$ or $x < -a$; we'll assume $x > a$ for the principal branch, and the constant $C$ will absorb sign adjustments.

2. Rewrite the square root.

   Compute $\frac{x + a}{x - a} = \frac{a\sec\theta + a}{a\sec\theta - a} = \frac{\sec\theta + 1}{\sec\theta - 1}$.

   Multiply numerator and denominator by $\sec\theta + 1$ to rationalise:

Hence $\sqrt{\frac{x + a}{x - a}} = \frac{|\sec\theta + 1|}{|\tan\theta|}$. For $\theta \in (0, \pi/2)$, $\sec\theta > 0$, $\tan\theta > 0$, so the absolute values drop:

3. Substitute into the integral. The integrand becomes:

Cancel $a$, $\sec\theta$, and $\tan\theta$:

4. Integrate. So the integral reduces to:

We know $\int \sec\theta \, d\theta = \log|\sec\theta + \tan\theta| + C$, and $\int 1 \, d\theta = \theta + C$.

Thus:

5. Back-substitute to $x$. Since $x = a \sec\theta$, we have $\sec\theta = \frac{x}{a}$. …

The key here is to notice that the denominator is a product of two expressions that look like $2\cosh x$ and $2\sinh x$ — but without the factor 2. Let’s rewrite:

So the integrand becomes:

Now recall the double-angle identity: $2\sinh x \cosh x = \sinh 2x$. Therefore:

So the integral is:

Now we need the antiderivative of $\operatorname{csch} u$. A standard result:

Let’s verify quickly: $\operatorname{csch} u = \frac{1}{\sinh u} = \frac{2}{e^u - e^{-u}}$. Substituting $t = e^u$ leads to a partial fraction, but the result is standard and worth memorising.

So with $u = 2x$, $du = 2\,dx$, we have:

Thus:

Now evaluate at the bounds. Let $a = \log\sqrt{2} = \frac{1}{2}\ln 2$, and $b = \log\sqrt{3} = \frac{1}{2}\ln 3$.

We need $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. At $x = \frac{1}{2}\ln k$, we have $e^x = \sqrt{k}$, $e^{-x} = 1/\sqrt{k}$. So:

$\tanh\left(\frac{1}{2}\ln k\right) = \frac{\sqrt{k} - 1/\sqrt{k}}{\sqrt{k} + 1/\sqrt{k}} = \frac{k - 1}{k + 1}.$ …

When faced with a complex integral, the first step is almost always to simplify the integrand. Here, we have terms involving $x^2$ and $(x \log x)^2$ under a square root. Our goal is to manipulate this expression into a recognizable standard form.

Notice the structure inside the square root: $4x^2 - (x \log x)^2$. Both terms contain $x^2$. This is a strong hint to factor out $x^2$.

Since the integration limits are from $1$ to $e$, $x$ is always positive, so $\sqrt{x^2} = x$.

Now the integral becomes:

The presence of $\log x$ and $\frac{1}{x}\, dx$ is a classic indicator for a substitution. Let $u = \log x$. Then $du = \frac{1}{x}\, dx$. This substitution will transform the integral into a much simpler form.

The resulting integral will be of the form $\int \frac{1}{\sqrt{a^2 - u^2}}\, du$. This is a standard integral that evaluates to $\arcsin\left(\frac{u}{a}\right)$. It's important to distinguish this from hyperbolic integrals. Hyperbolic inverse functions (like $\text{arsinh}$ or $\text{arccosh}$) arise from integrals involving $\sqrt{u^2 + a^2}$ or $\sqrt{u^2 - a^2}$, respectively. Our form $\sqrt{a^2 - u^2}$ is distinctly trigonometric.

Let's work through the steps:

1. Simplify the integrand: We begin by simplifying the expression under the square root.

Since $x$ is in the interval $[1, e]$, $x$ is positive. Therefore, $\sqrt{x^2} = x$.

Substituting this back into the integral, we get:

2. Perform a substitution:

   The structure of the integrand, with $\log x$ and $\frac{1}{x}\, dx$, strongly suggests the substitution $u = \log x$.

   Let $u = \log x$.

   Then, differentiating both sides with respect to $x$, we get $du = \frac{1}{x}\, dx$.

   We also need to change the limits of integration according to our substitution:

   • When $x = 1$, $u = \log 1 = 0$.
   • When $x = e$, $u = \log e = 1$.

3. Transform the integral:

   Substitute $u$ and $du$ into the integral with the new limits:

4. Recognize the standard integral form: …