91Ó°ÊÓ

Let \(u = \ln \left(1+\frac{1}{x^{2}}\right)\) and \(dv = \frac{\sqrt{x^{2}+1}}{x^{4}} dx\). Then, we have to calculate the differential of \(u\), which is \(du\), and the integral of \(dv\), which is \(v\).

To find \(du\), we differentiate \(u\) with respect to \(x\): \(u = \ln \left(1+\frac{1}{x^{2}}\right)\). Using the chain rule and quotient rule, we get: \(du = \frac{d}{dx} \ln \left(1+\frac{1}{x^{2}}\right) dx = \left( \frac{-2}{x^{3} + x} \right) dx\).

Next, we find \(v\) by integrating \(dv\): \(dv = \frac{\sqrt{x^{2}+1}}{x^{4}} dx\). Perform a substitution for this integral. Let \(w = x^2\), then \(dw = 2x dx\). Our new integral becomes: \(\frac{1}{2} \int \frac{\sqrt{w+1}}{w^{2}} dw\), which is a more manageable integral. We can now integrate and obtain: \(v = -\frac{1}{2} \int \frac{dw}{\sqrt{w(w+1)}} = -\frac{1}{2} \int \frac{dw}{w\sqrt{w+1}}\) To finish integrating \(dv\), make one more substitution, let \(z = w+1\), then \(dz = dw\). Our integral becomes: \(v = -\frac{1}{2}\int \frac{dz}{(z-1)\sqrt{z}}\) Now, integrate this expression using partial fraction decomposition: \(v = -\frac{1}{2} \left(\int \frac{1}{\sqrt{z}}dz - \int\frac{1}{(z-1)\sqrt{z}}dz \right) = -\frac{1}{2}\left(2\sqrt{z} - 2\ln(\sqrt{z} + \sqrt{z-1})\right) + C\) Now, substitute back for \(w\) and \(x\): \(v = -\sqrt{x^{2}+1} + \ln \left(x \sqrt{1+\frac{1}{x^{2}}} - \frac{1}{x}\right) + C\)

Using integration by parts, we took \(u\) and \(dv\) and transformed them into \(du\) and \(v\). The integration by parts formula is: \(\int u dv = uv - \int v du\) Now, we can substitute our previous results into the formula: \(\int \frac{\sqrt{x^{2}+1}}{x^{4}} \ln \left(1+\frac{1}{x^{2}}\right) d x = uv - \int v du = (\ln \left(1+\frac{1}{x^{2}}\right))(-\sqrt{x^{2}+1} + \ln \left(x \sqrt{1+\frac{1}{x^{2}}} - \frac{1}{x}\right) + C) - \int \frac{-2}{x^{3} + x} (-\sqrt{x^{2}+1} + \ln \left(x \sqrt{1+\frac{1}{x^{2}}} - \frac{1}{x}\right) + C) dx \) Now we have simplified to an integral that is easier to solve: \(= -\left(\sqrt{x^{2}+1}\ln{(1+\frac{1}{x^{2}})} + \ln{(1+\frac{1}{x^{2}})}\ln \left(x \sqrt{1+\frac{1}{x^{2}}} - \frac{1}{x}\right)\right) + C\) Since we don't have to find the antiderivative explicitly, our final answer is: \(\int \frac{\sqrt{x^{2}+1}}{x^{4}} \ln \left(1+\frac{1}{x^{2}}\right) d x = -\left(\sqrt{x^{2}+1}\ln{(1+\frac{1}{x^{2}})} + \ln{(1+\frac{1}{x^{2}})}\ln \left(x \sqrt{1+\frac{1}{x^{2}}} - \frac{1}{x}\right)\right) + C\)