Question

Evaluate the integral in terms of (a) natural logarithms and (b) inverse hyperbolic functions. \(\int_{-1 / 2}^{1 / 2} \frac{d x}{1-x^{2}}\)

Short answer

The result of the integral in terms of natural logarithms is \(\ln(3)\) and in terms of inverse hyperbolic functions, it is \(2\ln(3/2)\)

Step-by-step solution

Identifying the Integral Form

The integral \(\int_{-1 / 2}^{1 / 2} \frac{d x}{1-x^{2}}\) is an example of a basic algebraic fraction. It might be best to split it up into partial fractions, let's do: \(1/(1-x^{2}) = 1/(1-x) + 1/(1+x)\).

Integrate the split integral

We can now rewrite our integral as: \(\int_{-1/2}^{1/2} \frac{1}{1-x} dx + \int_{-1/2}^{1/2} \frac{1}{1+x} dx\). The antiderivative of \(1/(1-x)\) is \(-\ln|1-x|\) and of \(1/(1+x)\) is \(\ln|1+x|\), applying the fundamental theorem of calculus, we get: \(-[\ln(1/2) - \ln(3/2)] + [\ln(3/2) - \ln(1/2)]\).

Simplify the result

Simplify to get \( -\ln(1/3) + \ln(3)\), which can be rewritten using properties of logarithms as \( \ln(3^(3)/3) = \ln(3)\) for part (a).

Use hyperbolic functions

In terms of inverse hyperbolic functions, the function \(1/(1-x^2)\) can be recognised as the derivative of \(\text{artanh}(x)\), so the integral in part (b) can be rewritten as \(\text{artanh}(1/2) - \text{artanh}(-1/2)\), which leads to \(\ln(3/2) - (-\ln(3/2)) = 2\ln(3/2)\).