91Ó°ÊÓ

We will rewrite the integrand using partial fraction decomposition. Let $$ \frac{x^2}{x^4+a^4} = \frac{Ax+B}{x^2+\sqrt{2}ax} + \frac{Cx+D}{x^2-\sqrt{2}ax+a^2} $$ Multiplying both sides by the common denominator \((x^2+\sqrt{2}ax)(x^2-\sqrt{2}ax+a^2)\), we get $$ x^2 = (Ax + B)(x^2-\sqrt{2}ax+a^2) + (Cx + D)(x^2+\sqrt{2}ax). $$

We can now equate the coefficients on both sides of the equation, $$ x^2 = Ax^3 + (\sqrt{2}A)a x^2 + (a^2-\sqrt{2}aB) x + B(a^2) + \newline Cx^3 + (\sqrt{2}C)a x^2 - (a^2+\sqrt{2}aD) x + Da^2. $$ Grouping the terms by powers of \(x\), we obtain the following system of equations: $$ \begin{cases} 1 = \sqrt{2}A + \sqrt{2}C \\ 0 = A+C \\ 0 = a^2 -\sqrt{2}a B - \sqrt{2}a D \\ 0 = B(a^2)+Da^2 \end{cases} $$ Solving this system of equations, we find: $$ A = -\frac{1}{2\sqrt{2}}, \quad B = \frac{1}{2a^2}, \newline C = \frac{1}{2\sqrt{2}}, \quad D = \frac{1}{2a^2}. $$ Substituting these coefficients into the original partial fraction decomposition, we get $$ \frac{x^2}{x^4+a^4} = \frac{-\frac{1}{2\sqrt{2}}x+\frac{1}{2a^2}}{x^2+\sqrt{2}ax} + \frac{\frac{1}{2\sqrt{2}}x+\frac{1}{2a^2}}{x^2-\sqrt{2}ax+a^2}. $$

Now, we can integrate the rewritten integrand using substitution: $$ \int \frac{x^2dx}{x^4+a^4} = \int \left(\frac{-\frac{1}{2\sqrt{2}}x+\frac{1}{2a^2}}{x^2+\sqrt{2}ax} + \frac{\frac{1}{2\sqrt{2}}x+\frac{1}{2a^2}}{x^2-\sqrt{2}ax+a^2}\right)dx. $$ Let \(u=x^2+\sqrt{2}ax\) and \(v=x^2-\sqrt{2}ax+a^2\). Then \(du = (2x+\sqrt{2}a) dx\) and \(dv=(2x-\sqrt{2}a)dx\). Therefore: $$ \begin{aligned} \int\frac{-\frac{1}{2\sqrt{2}}x+\frac{1}{2a^2}}{x^2+\sqrt{2}ax}dx = -\frac{1}{2\sqrt{2}} \int\frac{u-1}{u}\frac{du}{2x+\sqrt{2}a}+\frac{1}{2a^2}\int\frac{1}{u}\frac{du}{2x+\sqrt{2}a} , \end{aligned} $$ and $$ \begin{aligned} \int\frac{\frac{1}{2\sqrt{2}}x+\frac{1}{2a^2}}{x^2-\sqrt{2}ax+a^2}dx = \frac{1}{2\sqrt{2}} \int\frac{v-a^2+1}{v}\frac{dv}{2x-\sqrt{2}a}+\frac{1}{2a^2}\int\frac{-1}{v}\frac{dv}{2x-\sqrt{2}a} . \end{aligned} $$ Now we notice that $$ \frac{du}{2x+\sqrt{2}a} = \frac{dv}{2x-\sqrt{2}a} = dx. $$ So, combining the two integrals gives us: $$ \begin{aligned} \int \frac{x^2 dx}{x^4+a^4} = -\frac{1}{2\sqrt{2}} \int\frac{u-1}{u}du+\frac{1}{2a^2}\int\frac{1}{u}du + \frac{1}{2\sqrt{2}} \int\frac{v-a^2+1}{v}dv+\frac{1}{2a^2}\int\frac{-1}{v}dv \end{aligned} $$ Using properties of logarithms, we can simplify and integrate: $$ \begin{aligned} \int \frac{x^2 dx}{x^4+a^4} =-\frac{1}{2\sqrt{2}}\left(\int{du} - \int \frac{1}{u}du\right) +\frac{1}{2a^2}\left(\int\frac{1}{u}du + \int\frac{-1}{v}dv\right) +\frac{1}{2\sqrt{2}}\left(\int{dv} - \int\frac{a^2-1}{v}dv\right) \end{aligned} $$ Now, integrate: $$ \begin{aligned} \int \frac{x^2 dx}{x^4+a^4} = -\frac{1}{2\sqrt{2}} (\ln |u|-\ln |x^2+\sqrt{2}ax|) +\frac{1}{2a^2} (\ln|u|-\ln|v|) +\frac{1}{2\sqrt{2}} (\ln|v|-\ln|(x^2-\sqrt{2}ax)+a^2|) \end{aligned} $$ Simplifying the expression and substituting \(u\) and \(v\) back, we get: $$ \int \frac{x^2 dx}{x^4+a^4} = -\frac{1}{2\sqrt{2}}\left(\ln \left|\frac{x^2+\sqrt{2}ax}{x^2}\right|\right) +\frac{1}{2a^2}\left(\ln\left|\frac{x^2+\sqrt{2}ax}{x^2-\sqrt{2}ax+a^2}\right|\right) + \newline \frac{1}{2\sqrt{2}}\left(\ln \left|\frac{x^2-\sqrt{2}ax+a^2}{x^2}\right|\right) + C, $$ where \(C\) is the constant of integration.