91Ó°ÊÓ

Since the denominator involves the term \((4-x^2)\), we can use the substitution: $$x = 2\sin u$$ This substitution simplifies the denominator. Also, notice that when \(u = 0\), \(x = 0\), which maintains the bounds of the integral if any.

Now we need to compute \(dx\) in terms of \(du\). Differentiating the substitution equation with respect to \(u\), we have: $$dx = 2\cos u du$$ Now we can substitute \(x = 2\sin u\) and \(dx = 2\cos u du\) back into the integral to transform it into an integral with respect to \(u\).

Substituting our expressions for \(x\) and \(dx\), we get: $$ \int \frac{(2\sin u)^2 (2\cos u) du}{\left(4-(2\sin u)^2\right)^{5/2}} $$

Now, we can simplify the integral further: $$ \int \frac{4\sin^2(u) 2\cos(u) du}{\left(4-4\sin^2(u)\right)^{5/2}} $$ Use the trigonometric identity \(\cos^2(u) = 1- \sin^2(u)\) to rewrite the denominator of the integrand: $$ \int \frac{4\sin^2(u) 2\cos(u) du}{(4\cos^2(u))^{5/2}} $$ Simplify the expression: $$ \int \frac{8\sin^2(u)\cos(u) du}{8\cos^5(u)} $$ Reduce the expression to the following integral: $$ \int \frac{\sin^2(u)}{\cos^4(u)} \cos(u) du $$ Use the trigonometric identity \(\sin^2(u) = 1- \cos^2(u)\) to rewrite the integrand: $$ \int \frac{1 - \cos^2(u)}{\cos^4(u)} \cos(u) du $$ Separate the integral into two parts: $$ \int \left(\frac{1}{\cos^3(u)} - \frac{\cos(u)}{\cos^5(u)}\right) du $$ Now integrate each part separately: $$ \int\sec^3(u) du - \int\sec^5(u) d(u) $$ Compute the integrals by using the standard methods for integrating secant functions: $$ \frac{1}{2}(\sec(u)\tan(u) + \ln|\sec(u) + \tan(u)|) - (\frac{3}{8}\tan(u)\sec(u) + \frac{1}{8}\ln|\sec(u) + \tan(u)|) + C $$ We have now integrated in terms of \(u\). We should return the result to the original variable: \(x\).

Use the substitution \(x = 2\sin u\) to find the inverse function: $$ u = \arcsin\left(\frac{x}{2}\right) $$ Substitute the expressions for \(u\) and calculate the corresponding trigonometric functions: $$ \tan(u) = \frac{x}{\sqrt{4-x^2}}, \quad \sec(u) =\frac{2}{\sqrt{4-x^2}} $$ Now, plug these values back into the integral result: $$ \frac{1}{2}\left(\frac{2x}{4-x^2} + \ln\left|\frac{2}{\sqrt{4-x^2}} + \frac{x}{\sqrt{4-x^2}}\right|\right) - \left(\frac{3}{8}\frac{x}{\sqrt{4-x^2}}\frac{2}{\sqrt{4-x^2}} + \frac{1}{8}\ln\left|\frac{2}{\sqrt{4-x^2}} + \frac{x}{\sqrt{4-x^2}}\right|\right) + C $$ Combining the log terms and simplifying: $$ \frac{1}{8}\left(8\frac{x}{4-x^2}+\ln\left|\frac{2+x}{\sqrt{4-x^2}}\right|\right)-\frac{3}{4}\cdot\frac{x^2}{4-x^2} + C $$ And that's our final result: $$ \int \frac{x^2 dx}{(4-x^2)^{5/2}}=\frac{1}{8}\left(8\frac{x}{4-x^2}+\ln\left|\frac{2+x}{\sqrt{4-x^2}}\right|\right)-\frac{3}{4}\cdot\frac{x^2}{4-x^2} + C $$