Question

Find the integral. $$ \int \frac{1}{x \sqrt{4 x^{2}-1}} d x $$

Short answer

The integral of \(\int \frac{1}{x \sqrt{4 x^{2}-1}} d x\) is \(C + sec^{-1}(2x)\)

Step-by-step solution

Identify an appropriate substitution

We can rewrite \(4x^2 - 1\) as \((2x)^2 - 1^2\), which resembles the derivative of arc secant (\(sec^{-1}\)). Hence, we perform a substitution with \(2x = sec(u)\) or \(x = \frac{sec(u)}{2}\). This yields \(dx = \frac{sec(u)tan(u)}{2} du\). We substitute these into the original integral.

Substitute the values into the integral

Substituting into the integral, we get:\[\int \frac{1}{\frac{sec(u)}{2}\sqrt{4(\frac{sec(u)}{2})^{2}-1}} \cdot \frac{sec(u)tan(u)}{2} du\]This simplifies to:\[\int \frac{sec(u)tan(u)}{sec(u)\sqrt{sec^2(u)-1}} du\]As tangent is the ratio of sine to cosine, and secant is the reciprocal of cosine, we can see that the integral simplifies even further to:\[\int \frac{sec(u)tan(u)}{sec(u)\sqrt{sec^2(u)-1}} du = \int \frac{tan(u)}{\sqrt{sec^2(u)-1}} du\]This is equal to: \(\int du\).

Evaluate the Integral

The integral of \(du\) is simply \(u + C\), where \(C\) is the constant of integration.

Substitute the original value back

Substituting the original value back into the equation, we replace \(u\) with \(sec^{-1}(2x)\) to get the final answer:C + \(sec^{-1}(2x)\).