Problem 10 - Essential Calculus

Question

Find the integral. $$ \int \frac{1}{x \sqrt{x^{4}-4}} d x $$

Step-by-Step Solution

Answer

The evaluated integral is \(\frac{1}{2}\ln|x^{2}/\sqrt{x^{4}-4}+ \sqrt{x^{4}/4-1}|\) + C.

Step 1: Perform a substitution

Let's perform a substitution to simplify this integral. Set \(u = x^2\). This means \(du = 2x \, dx\), and in terms of \(x \, dx\) is \(du/2\). Thus, the original integral now becomes \(\int \frac{du}{2u \sqrt{u^{2}-4}}\).

Step 2: Further simplify the integral

We can simplify the integral by taking out the constant term from the integral and applying the rule \( \int f(x) \, dx = F(x) + C \), which states that the integral of a function is its antiderivative plus a constant. This gives us: \(\frac{1}{2}\int \frac{du}{u \sqrt{u^{2}-4}}\).

Step 3: Use a standard integral formula

This integral is now in the form of a standard integral: \(\int \frac{du}{u \sqrt{u^{2}-a^{2}}}\) where \(a = 2\). The standard integral has the result: \(\frac{1}{a}\ln|u/\sqrt{u^{2}-a^{2}}+ \sqrt{u^{2}/a^{2}-1}|+C\). Therefore, substituting \(a = 2\), the result is \(\frac{1}{2}\ln|u/\sqrt{u^{2}-4}+ \sqrt{u^{2}/4-1}|\).

Step 4: Substitute back

Substitute \(u = x^2\) back, the result is then \(\frac{1}{2}\ln|x^{2}/\sqrt{x^{4}-4}+ \sqrt{x^{4}/4-1}|\).