We are asked to evaluate the following three definite integrals: $I = \int_{1}^{4} (\frac{2}{\sqrt{x}} + 6x^2 - 5) dx$ $J = \int_{0}^{\frac{\pi}{4}} (1 - 2\sin^2 x) dx$ $K = \int_{2}^{3} \frac{x^3 + x^2 - 1}{x^2 - 1} dx$

1. Problem Description

We are asked to evaluate the following three definite integrals:

$I = \int_{1}^{4} (\frac{2}{\sqrt{x}} + 6x^2 - 5) dx$

$J = \int_{0}^{\frac{\pi}{4}} (1 - 2\sin^2 x) dx$

$K = \int_{2}^{3} \frac{x^3 + x^2 - 1}{x^2 - 1} dx$

2. Solution Steps

For $I$:

$I = \int_{1}^{4} (\frac{2}{\sqrt{x}} + 6x^2 - 5) dx = \int_{1}^{4} (2x^{-1/2} + 6x^2 - 5) dx$

$I = [2 \cdot \frac{x^{1/2}}{1/2} + 6 \cdot \frac{x^3}{3} - 5x]_1^4 = [4\sqrt{x} + 2x^3 - 5x]_1^4$

$I = (4\sqrt{4} + 2(4^3) - 5(4)) - (4\sqrt{1} + 2(1^3) - 5(1)) = (4(2) + 2(64) - 20) - (4 + 2 - 5)$

$I = (8 + 128 - 20) - (1) = 116 - 1 = 115$

For $J$:

$J = \int_{0}^{\frac{\pi}{4}} (1 - 2\sin^2 x) dx$

Using the trigonometric identity $\cos(2x) = 1 - 2\sin^2(x)$, we have:

$J = \int_{0}^{\frac{\pi}{4}} \cos(2x) dx$

$J = [\frac{1}{2} \sin(2x)]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \sin(2 \cdot \frac{\pi}{4}) - \frac{1}{2} \sin(0) = \frac{1}{2} \sin(\frac{\pi}{2}) - 0$

$J = \frac{1}{2} (1) = \frac{1}{2}$

For $K$:

$K = \int_{2}^{3} \frac{x^3 + x^2 - 1}{x^2 - 1} dx$

We perform polynomial long division:

$\frac{x^3 + x^2 - 1}{x^2 - 1} = x + 1 + \frac{x}{x^2 - 1}$

Thus, $K = \int_{2}^{3} (x + 1 + \frac{x}{x^2 - 1}) dx$

$K = \int_{2}^{3} (x + 1) dx + \int_{2}^{3} \frac{x}{x^2 - 1} dx$

$K = [\frac{x^2}{2} + x]_2^3 + \frac{1}{2} \int_{2}^{3} \frac{2x}{x^2 - 1} dx$

$K = (\frac{3^2}{2} + 3) - (\frac{2^2}{2} + 2) + \frac{1}{2} [\ln |x^2 - 1|]_2^3$

$K = (\frac{9}{2} + 3) - (2 + 2) + \frac{1}{2} (\ln |3^2 - 1| - \ln |2^2 - 1|) = (\frac{15}{2}) - 4 + \frac{1}{2} (\ln 8 - \ln 3)$

$K = \frac{15}{2} - \frac{8}{2} + \frac{1}{2} \ln(\frac{8}{3}) = \frac{7}{2} + \frac{1}{2} \ln(\frac{8}{3})$

$K = \frac{7}{2} + \frac{1}{2} \ln(\frac{8}{3})$

3. Final Answer

$I = 115$

$J = \frac{1}{2}$

$K = \frac{7}{2} + \frac{1}{2}\ln(\frac{8}{3})$