Calculate the following indefinite integral.

[1] $\int \frac{e^{2x}}{(e^x+1)^2}dx$

[2] $\int \sin x\cos 3x dx$

[3] $\int \sin 2x\sin 3x dx$

[4] $\int \frac{dx}{4x^2-12x+9}$

[5] $\int \cos ^4 x dx$
1. **Integral 1:**

   \[
   \int \frac{e^{2x}}{(e^x+1)^2} \, dx
   \]

   Let \( u = e^x + 1 \). Then \( du = e^x \, dx \) and \( e^x = u - 1 \).

   Substituting these into the integral, we get:

   \[
   \int \frac{e^{2x}}{(e^x+1)^2} \, dx = \int \frac{(u-1)^2}{u^2} \cdot \frac{du}{u-1} = \int \frac{u^2 - 2u + 1}{u^2} \cdot \frac{du}{u-1}
   \]

   Simplifying the integrand:

   \[
   \int \left( \frac{u^2}{u^2} - \frac{2u}{u^2} + \frac{1}{u^2} \right) \, du = \int \left( 1 - \frac{2}{u} + \frac{1}{u^2} \right) \, du
   \]

   Integrating term by term:

   \[
   \int 1 \, du - 2 \int \frac{1}{u} \, du + \int \frac{1}{u^2} \, du = u - 2 \ln |u| - \frac{1}{u} + C
   \]

   Substituting back \( u = e^x + 1 \):

   \[
   e^x + 1 - 2 \ln |e^x + 1| - \frac{1}{e^x + 1} + C
   \]

   Simplifying the expression:

   \[
   \boxed{\ln (e^x+1) + \frac{1}{e^x+1} + C}
   \]

2. **Integral 2:**

   \[
   \int \sin x \cos 3x \, dx
   \]

   Using the product-to-sum identities:

   \[
   \sin \alpha \cos \beta = \frac{1}{2} \left( \sin (\alpha + \beta) + \sin (\alpha - \beta) \right)
   \]

   Let \(\alpha = x\) and \(\beta = 3x\):

   \[
   \sin x \cos 3x = \frac{1}{2} \left( \sin (4x) + \sin (-2x) \right) = \frac{1}{2} \left( \sin 4x - \sin 2x \right)
   \]

   Therefore:

   \[
   \int \sin x \cos 3x \, dx = \frac{1}{2} \int \left( \sin 4x - \sin 2x \right) \, dx
   \]

   Integrating term by term:

   \[
   \frac{1}{2} \left( -\frac{1}{4} \cos 4x + \frac{1}{2} \cos 2x \right) + C
   \]

   Simplifying:

   \[
   \boxed{\frac{1}{4} \cos 2x - \frac{1}{8} \cos 4x + C}
   \]

3. **Integral 3:**

   \[
   \int \sin 2x \sin 3x \, dx
   \]

   Using the product-to-sum identities:

   \[
   \sin \alpha \sin \beta = \frac{1}{2} \left( \cos (\alpha - \beta) - \cos (\alpha + \beta) \right)
   \]

   Let \(\alpha = 2x\) and \(\beta = 3x\):

   \[
   \sin 2x \sin 3x = \frac{1}{2} \left( \cos (2x - 3x) - \cos (2x + 3x) \right) = \frac{1}{2} \left( \cos (-x) - \cos 5x \right)
   \]

   Since \(\cos (-x) = \cos x\):

   \[
   \sin 2x \sin 3x = \frac{1}{2} \left( \cos x - \cos 5x \right)
   \]

   Therefore:

   \[
   \int \sin 2x \sin 3x \, dx = \frac{1}{2} \int \left( \cos x - \cos 5x \right) \, dx
   \]

   Integrating term by term:

   \[
   \frac{1}{2} \left( \sin x - \frac{1}{5} \sin 5x \right) + C
   \]

   Simplifying:

   \[
   \boxed{\frac{1}{2} \sin x - \frac{1}{10} \sin 5x + C}
   \]

4. **Integral 4:**

   \[
   \int \frac{dx}{4x^2 - 12x + 9}
   \]

   Completing the square in the denominator:

   \[
   4x^2 - 12x + 9 = (2x - 3)^2
   \]

   Therefore:

   \[
   \int \frac{dx}{(2x - 3)^2}
   \]

   Let \( u = 2x - 3 \). Then \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \).

   Substituting these into the integral:

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

   Integrating:

   \[
   \frac{1}{2} \left( -u^{-1} \right) + C = -\frac{1}{2u} + C
   \]

   Substituting back \( u = 2x - 3 \):

   \[
   \boxed{-\frac{1}{2(2x - 3)} + C}
   \]

5. **Integral 5:**

   \[
   \int \cos^4 x \, dx
   \]

   Using the power-reduction formula:

   \[
   \cos^2 \theta = \frac{1 + \cos 2\theta}{2}
   \]

   Therefore:

   \[
   \cos^4 x = \left( \cos^2 x \right)^2 = \left( \frac{1 + \cos 2x}{2} \right)^2
   \]

   Expanding the square:

   \[
   \cos^4 x = \frac{1}{4} \left( 1 + 2 \cos 2x + \cos^2 2x \right)
   \]

   Using the power-reduction formula again for \(\cos^2 2x\):

   \[
   \cos^2 2x = \frac{1 + \cos 4x}{2}
   \]

   Substituting this back:

   \[
   \cos^4 x = \frac{1}{4} \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) = \frac{1}{4} \left( \frac{3}{2} + 2 \cos 2x + \frac{\cos 4x}{2} \right)
   \]

   Therefore:

   \[
   \int \cos^4 x \, dx = \frac{1}{4} \int \left( \frac{3}{2} + 2 \cos 2x + \frac{\cos 4x}{2} \right) \, dx
   \]

   Integrating term by term:

   \[
   \frac{1}{4} \left( \frac{3}{2} x + \sin 2x + \frac{1}{8} \sin 4x \right) + C
   \]

   Simplifying:

   \[
   \boxed{\frac{3}{8} x + \frac{1}{4} \sin 2x + \frac{1}{32} \sin 4x + C}
   \]