(a) To find \(\int_{1}^{\infty} f(x) \, dx\), we first integrate \(f(x) = \frac{2}{(x+2)^2}\).

The integral of \(\frac{2}{(x+2)^2}\) is \(\frac{-2}{x+2}\).

Evaluate the definite integral from 1 to \(\infty\):

\(\left[ \frac{-2}{x+2} \right]_1^{\infty} = 0 - \left( \frac{-2}{3} \right) = \frac{2}{3}\).

(b) Given \(\frac{dy}{dx} = \frac{2}{(x+2)^2}\), integrate to find \(y\):

\(y = \frac{-2}{x+2} + c\).

Using the point \((-1, -1)\), substitute \(x = -1\) and \(y = -1\):

\(-1 = \frac{-2}{-1+2} + c\).

\(-1 = -2 + c\).

\(c = 1\).

Thus, the equation of the curve is \(y = \frac{-2}{x+2} + 1\).

(a) To find the equation of the normal, first find the gradient of the tangent at \(x = 2\) using \(\frac{dy}{dx} = \frac{1}{2}x + \frac{72}{x^4}\).

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{1}{2}(2) + \frac{72}{2^4} = 1 + \frac{72}{16} = 1 + 4.5 = 5.5\).

The gradient of the normal is the negative reciprocal: \(-\frac{1}{5.5} = -\frac{2}{11}\).

Using point-slope form \(y - y_1 = m(x - x_1)\) with \(P(2, 8)\):

\(y - 8 = -\frac{2}{11}(x - 2)\).

Simplifying gives \(y = -\frac{2}{11}x + \frac{92}{11}\).

(b) Integrate \(\frac{dy}{dx} = \frac{1}{2}x + \frac{72}{x^4}\) to find \(y\):

\(y = \int \left( \frac{1}{2}x + \frac{72}{x^4} \right) dx = \frac{1}{4}x^2 - \frac{24}{x^3} + C\).

Use the point \(P(2, 8)\) to find \(C\):

\(8 = \frac{1}{4}(2)^2 - \frac{24}{2^3} + C\).

\(8 = 1 - 3 + C\).

\(C = 10\).

Thus, the equation of the curve is \(y = \frac{1}{4}x^2 - \frac{24}{x^3} + 10\).

To find the equation of the curve, we need to integrate the derivative \(f'(x) = 6x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}\).

Integrating term by term:

\(\int 6x^{-\frac{1}{2}} \, dx = 6 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 12x^{\frac{1}{2}}\)

\(\int -4x^{-\frac{3}{2}} \, dx = -4 \cdot \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} = 8x^{-\frac{1}{2}}\)

Thus, the integral of \(f'(x)\) is:

\(f(x) = 12x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} + c\)

We use the point (4, 7) to find \(c\):

\(7 = 12(4)^{\frac{1}{2}} + 8(4)^{-\frac{1}{2}} + c\)

\(7 = 12 \cdot 2 + 8 \cdot \frac{1}{2} + c\)

\(7 = 24 + 4 + c\)

\(c = 7 - 28 = -21\)

Therefore, the equation of the curve is:

\(y = 12x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} - 21\)

To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = \frac{1}{(x-3)^2} + x\).

First, integrate \(\frac{1}{(x-3)^2}\):

\(\int \frac{1}{(x-3)^2} \, dx = -(x-3)^{-1} + C_1\)

Next, integrate \(x\):

\(\int x \, dx = \frac{1}{2}x^2 + C_2\)

Combine the results of the integration:

\(y = -(x-3)^{-1} + \frac{1}{2}x^2 + c\)

Use the point (2, 7) to find the constant \(c\):

\(7 = -\frac{1}{(2-3)} + \frac{1}{2}(2)^2 + c\)

\(7 = 1 + 2 + c\)

\(c = 4\)

Thus, the equation of the curve is:

\(y = -(x-3)^{-1} + \frac{1}{2}x^2 + 4\)

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}\).

Integrating term by term:

\(y = \int (3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}) \, dx\)

\(= \int 3x^{\frac{1}{2}} \, dx - \int 3x^{-\frac{1}{2}} \, dx\)

\(= \frac{3}{\frac{3}{2}} x^{\frac{3}{2}} - \frac{3}{\frac{1}{2}} x^{\frac{1}{2}} + c\)

\(= 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + c\)

Using the point (4, 7) to find \(c\):

\(7 = 2(4)^{\frac{3}{2}} - 6(4)^{\frac{1}{2}} + c\)

\(7 = 16 - 12 + c\)

\(7 = 4 + c\)

\(c = 3\)

Thus, the equation of the curve is \(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 3\).