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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

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