91Ó°ÊÓ

Let's perform the substitution: $$x = 2\sinh{u}$$ The differential of x is: $$dx = 2\cosh{u} du$$ Now, we'll replace \(x\) and \(dx\) in the original integral: $$\int \frac{(2\sinh{u})^2}{\sqrt{4+(2\sinh{u})^2}} (2\cosh{u}) du$$

Let's simplify the expression in the integral: $$\int \frac{4\sinh^2{u}}{\sqrt{4+4\sinh^2{u}}} (2\cosh{u}) du$$ $$\int \frac{4\sinh^2{u}\cosh{u}}{2\sqrt{1+\sinh^2{u}}} du$$ $$\int 2\sinh^2{u}\cosh{u}\,du$$ Since \(\cosh^2{u}-\sinh^2{u}=1\), we can rewrite the integral as: $$\int 2(\cosh^2{u}-1)\cosh{u}\,du$$

We can split the integral into two separate integrals and then evaluate each of them: $$2\int \cosh^3{u}\,du - 2\int \cosh{u}\,du$$ For the first integral, we can use the reduction formula for \(\cosh^n{u}\), which is: $$\int \cosh^n{u}\,du = \frac{1}{n} \cosh^{n-1}{u} \sinh{u} + \frac{n-1}{n}\int \cosh^{n-2}{u}\,du$$ Applying this to our integral, we have: $$\int \cosh^3{u} \,du = \frac{1}{3}\cosh^2{u}\sinh{u} + \frac{2}{3}\int \cosh{u}\,du$$ Finally, the second integral is easy to compute as we know that the derivative of \(\sinh{u}\) is \(\cosh{u}\). So, the result of the first integral is: $$\frac{1}{3}\cosh^2{u}\sinh{u} + \frac{4}{3}\sinh{u}$$ Substitute this back into our original expression, we have: $$2\left(\frac{1}{3}\cosh^2{u}\sinh{u} + \frac{4}{3}\sinh{u}\right) - 2\sinh{u}$$ $$\frac{2}{3}\cosh^2{u}\sinh{u} + \frac{4}{3}\sinh{u}$$

Let's substitute back \(x = 2\sinh{u}\) to get the result in terms of x: $$\frac{2}{3}(1+\sinh^2{u})\sinh{u} + \frac{4}{3}\sinh{u}$$ $$\frac{2}{3}(1+\frac{x^2}{4})\frac{x}{2} + \frac{4}{3}\frac{x}{2}$$ $$\frac{1}{3}(x^2+4)x$$ So, the integral is: $$\int \frac{x^{2}}{\sqrt{4+x^{2}}} dx = \frac{1}{3}(x^2+4)x + C$$ where C is the constant of integration.