Question

Evaluate \(\int_{1}^{1 / 2} \frac{x^{2}+1}{x^{4}-x^{2}+1} \ln \left(1+x-\frac{1}{x}\right) d x\)

Short answer

Using the above step-by-step solution, provide a short answer: The given integral can be simplified to \(\int_{1}^{1 / 2} \frac{x^{2}+1}{x^{4}-x^{2}+1} \ln\left(1+x-\frac{1}{x}\right) d x = \int_{1}^{1 / 2} \frac{x^2 +1}{x^4 - x^2 + 1} \ln \left(x^2 + x^2 - 1\right) dx\) . In order to evaluate this integral, numerical methods such as Simpson's rule would be required.

Step-by-step solution

Integrand Simplification

First, let's focus on the logarithm part of the integrand. We see that it is: \( \ln \left(1+x-\frac{1}{x}\right)\). Multiply both the numerator and denominator inside the logarithm by x to simplify this term: \( \ln \left(\frac{x^2 + x^2 - 1}{x}\right) = \ln \left(x^2 + x^2 - 1\right) - \ln(x)\) Now substitute this back into the integrand: \( \frac{x^2 +1}{x^4 - x^2 + 1} \ln \left(1+x-\frac{1}{x}\right) dx = \frac{x^2 +1}{x^4 - x^2 + 1} (\ln \left(x^2 + x^2 - 1\right) - \ln(x)) dx\) Now we can split the integrand into two parts: \( \frac{x^2 +1}{x^4 - x^2 + 1} \ln \left(x^2 + x^2 - 1\right) dx - \frac{x^2 +1}{x^4 - x^2 + 1} \ln(x) dx\)

Integration by Parts

We can keep the first part as it is: \(I_1 = \int_{1}^{1 / 2} \frac{x^2 +1}{x^4 - x^2 + 1} \ln \left(x^2 + x^2 - 1\right) dx\) Since the integrand in the second part has both a logarithm and a rational function, we can try integration by parts, using this formula: \(\int u\,dv = uv - \int v\,du\). Let's choose \(u = \ln(x)\) and \(dv = \frac{x^2 +1}{x^4 - x^2 + 1} dx\). Now, we need to calculate \(du\) and \(v\). For \(du\), differentiate \(u\) with respect to \(x\), and for \(v\), integrate \(dv\) with respect to \(x\): \(du = \frac{1}{x} dx\) \(v = \int \frac{x^2 +1}{x^4 - x^2 + 1} dx\) The integration for \(v\) is not straightforward. However, we can notice that the denominator is symmetric \((x^4 - x^2 + 1 = (1-x^2)+x^4)\), and the limits of integration are also symmetric around 1. Therefore, the integral evaluates to 0, so we can say: \(v = 0\) Now substitute these values back into the integration by parts formula: \(I_2=-\int_{1}^{1 / 2} \frac{x^2 +1}{x^4 - x^2 + 1} \ln(x) dx = 0 - 0 = 0\)

Final Evaluation

Now we have simplified the integral into two parts where the second part evaluates to 0. Therefore, the integral can be written as: \(\int_{1}^{1 / 2} \frac{x^{2}+1}{x^{4}-x^{2}+1} \ln\left(1+x-\frac{1}{x}\right) d x = I_1 + I_2\) \(I_1 = \int_{1}^{1 / 2} \frac{x^2 +1}{x^4 - x^2 + 1} \ln \left(x^2 + x^2 - 1\right) dx\) Now, we need to numerically evaluate the definite integral \(I_1\). However, this integral likely cannot be evaluated using elementary functions, and a numerical method (such as Simpson's rule) would be required. In conclusion, the given integral can be simplified to: \(\int_{1}^{1 / 2} \frac{x^{2}+1}{x^{4}-x^{2}+1} \ln\left(1+x-\frac{1}{x}\right) d x = \int_{1}^{1 / 2} \frac{x^2 +1}{x^4 - x^2 + 1} \ln \left(x^2 + x^2 - 1\right) dx\) And its evaluation would require the use of numerical methods.