91Ó°ÊÓ

Let's choose the substitution: $$ u = x^2 \Rightarrow x = \sqrt{u} $$ Now we'll differentiate our substitution in terms of x to find the needed dx.

Differentiating both sides of the substitution with respect to x, we get: $$ \frac{du}{dx} = 2x \Rightarrow dx = \frac{du}{2x} $$

Now we'll substitute the expressions for \(x^2\), \(x\), and \(dx\) from step 1 and 2: $$ \int \frac{(1-u) \frac{du}{2\sqrt{u}}}{\sqrt{u} (1+u)^{3 / 2}} $$ Let's simplify the integral: $$ \frac{1}{2} \int \frac{(1-u)du}{u(1+u)^{3/2} } $$

We'll use the method of partial fractions to decompose the integrand. The general form of the decomposition is: $$ \frac{1-u}{u(1+u)^{3/2}} = \frac{A}{\sqrt{u}} + \frac{B}{(1+u)^{1/2}} + \frac{C}{(1+u)^{3/2}} $$ Multiplying both sides by \(u(1+u)^{3/2}\), we get: $$ 1-u = A(1+u)^{3/2} + B u (1+u) + Cu(1+u)^{1/2}. $$ By setting \(u=0\), we obtain \(A=-1\). Then, the equation becomes: $$ 1-u = -\sqrt{1+u} + B u(1+u) + C u\sqrt{1+u}. $$ Now, we will differentiate both sides with respect to u: $$ -1 = \frac{1}{2\sqrt{1+u}} + B(1+2u) + \frac{C}{2\sqrt{1+u}} + C\sqrt{1+u}. $$ Setting \(u=-1\), we get \(B=-\frac{1}{2}\). Finally, setting \(u=0\), we obtain \(C=-\frac{1}{2}\). The integral becomes: $$ \frac{1}{2}\int\left(-\frac{1}{\sqrt{u}} - \frac{u}{2(1+u)^{1/2}}- \frac{u}{2(1+u)^{3/2}} \right) du $$ Now we can integrate each term separately: $$ \frac{1}{2}\left(-2\sqrt{u} - u\sqrt{1+u} - \frac{1}{\sqrt{1+u}}\right) + c = -\sqrt{u} -\frac{u\sqrt{1+u}}{2} - \frac{1}{2\sqrt{1+u}} + c $$

Finally, we'll substitute back the original expression for \(u\) from step 1: $$ - \sqrt{x^2} + \frac{x^2\sqrt{1+x^2}}{2} + \frac{1}{2\sqrt{1+x^{2}}} + c = -x + \frac{x^2\sqrt{1+x^2}}{2} + \frac{1}{2\sqrt{1+x^{2}}} + c $$ So the evaluated integral is: $$ \int \frac{\left(1-x^{2}\right) d x}{x^{1 / 2}\left(1+x^{2}\right)^{3 / 2}} = -x + \frac{x^2\sqrt{1+x^2}}{2} + \frac{1}{2\sqrt{1+x^{2}}} + c $$