Let's solve each of the integrals given in the image one by one.

1. $\int_1^4 \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx$

   Let $u = \sqrt{x}$. Then $du = \frac{1}{2\sqrt{x}} dx$ or $dx = 2\sqrt{x} \, du = 2u \, du$. When $x = 1$, $u = 1$. When $x = 4$, $u = 2$.

   The integral becomes: $\int_1^4 \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx = \int_1^2 \frac{e^u}{u} \cdot 2u \, du = 2 \int_1^2 e^u \, du$ $2 \int_1^2 e^u \, du = 2 \left[ e^u \right]_1^2 = 2(e^2 - e^1) = 2(e^2 - e)$

2. $\int_1^e \ln^2 x \, dx$

   Let $u = \ln x$. Then $du = \frac{1}{x} dx$ or $dx = x \, du$. When $x = 1$, $u = 0$. When $x = e$, $u = 1$.

   The integral becomes: $\int_1^e \ln^2 x \, dx = \int_0^1 u^2 e^u \, du$

   Using integration by parts, let $v = u^2$ and $dw = e^u \, du$. Then, $dv = 2u \, du$ and $w = e^u$. $\int u^2 e^u \, du = u^2 e^u - \int 2u e^u \, du$ The remaining integral again requires integration by parts. Let $v = 2u$ and $dw = e^u \, du$. Then, $dv = 2 \, du$ and $w = e^u$. $\int 2u e^u \, du = 2u e^u - \int 2 e^u \, du = 2u e^u - 2e^u$ So, $\int u^2 e^u \, du = u^2 e^u - (2u e^u - 2e^u) = u^2 e^u - 2u e^u + 2e^u$ Evaluating from 0 to 1: $\left[ u^2 e^u - 2u e^u + 2e^u \right]_0^1 = (1^2 e^1 - 2 \cdot 1 \cdot e^1 + 2e^1) - (0^2 e^0 - 2 \cdot 0 \cdot e^0 + 2e^0)$ $= (e - 2e + 2e) - (0 - 0 + 2) = e - e + 2 = e + 2 - 2 = e - 1$

3. $\int_0^3 \frac{x^2 \, dx}{\sqrt{9 - x^2}}$

   Let $x = 3 \sin t$. Then $dx = 3 \cos t \, dt$. When $x = 0$, $t = 0$. When $x = 3$, $t = \frac{\pi}{2}$.

   The integral becomes: $\int_0^3 \frac{x^2 \, dx}{\sqrt{9 - x^2}} = \int_0^{\pi/2} \frac{(3 \sin t)^2 \cdot 3 \cos t \, dt}{\sqrt{9 - 9 \sin^2 t}} = 27 \int_0^{\pi/2} \frac{\sin^2 t \cos t \, dt}{\sqrt{9 \cos^2 t}}$ $= 27 \int_0^{\pi/2} \frac{\sin^2 t \cos t \, dt}{3 \cos t} = 9 \int_0^{\pi/2} \sin^2 t \, dt$ Using the identity $\sin^2 t = \frac{1 - \cos(2t)}{2}$, $= 9 \int_0^{\pi/2} \frac{1 - \cos(2t)}{2} \, dt = \frac{9}{2} \left[ t - \frac{\sin(2t)}{2} \right]_0^{\pi/2}$ $= \frac{9}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{9\pi}{4}$

4. $\int_{\pi/4}^{\pi/2} \frac{\cos^4 x \, dx}{\sin^6 x}$

   Rewrite