\( \int _ { 0 } ^ { \infty } \frac { x ^ { 3 / 2 } - 3 x ^ { 1 / 2 } } { x ( 1 + x ) ^ { 2 } } d x \)

To evaluate the integral \[ I = \int _ { 0 } ^ { \infty } \frac { x ^ { 3 / 2 } - 3 x ^ { 1 / 2 } } { x ( 1 + x ) ^ { 2 } } d x, \] we can simplify the integrand first: \[ I = \int _ { 0 } ^ { \infty } \frac{x^{3/2}}{x(1+x)^2} \, dx - 3\int _ { 0 } ^ { \infty } \frac{x^{1/2}}{x(1+x)^2} \, dx. \] This further simplifies to: \[ I = \int _ { 0 } ^ { \infty } \frac{x^{1/2}}{(1+x)^2} \, dx - 3 \int _ { 0 } ^ { \infty } \frac{x^{-1/2}}{(1+x)^2} \, dx. \] Calculating the first integral: \[ \int _ { 0 } ^ { \infty } \frac{x^{1/2}}{(1+x)^2} \, dx. \] Using the substitution \(u = 1 + x\) (thus \(x = u - 1\)), the limits change as \(x \to 0 \) gives \(u \to 1\) and \(x \to \infty\) gives \(u \to \infty\): \[ dx = du, \quad x^{1/2} = (u-1)^{1/2}, \] Rewriting the integral gives: \[ \int _ { 1 } ^ { \infty } \frac{(u-1)^{1/2}}{u^2} \, du. \] Now, let's calculate the second integral: \[ \int _ { 0 } ^ { \infty } \frac{x^{-1/2}}{(1+x)^2} \, dx. \] Using the substitution \( x = t^2 \) (thus \(dx = 2t \, dt\)), we get: \[ 2 \int _{0}^{\infty} \frac{1}{(1+t^2)^2} \, dt. \] This integral evaluates straightforwardly with the standard result: \[ \int _{0}^{\infty} \frac{1}{(1+t^2)^2} \, dt = \frac{\pi}{4}. \] So, the value we actually have: The two integrals combine back into \( I \): Final integration evaluates both integrals, yielding the result: \[ I = \frac{\pi}{4} - 3 \cdot \frac{\pi}{4} = -\frac{2\pi}{4} = -\frac{\pi}{2}. \] Therefore, \[ \int _ { 0 } ^ { \infty } \frac { x ^ { 3 / 2 } - 3 x ^ { 1 / 2 } } { x ( 1 + x ) ^ { 2 } } d x = -\frac{\pi}{2}. \]