Calculate the following indefinite integrals.

[1] $\int \sin x\cos ^ 3 x dx$

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

[3] $\int x^2 \sqrt{x^3+1}dx$

[4] $\int \frac{e^{2x}-3e^{x}}{e^x}dx$

[5] $\int (1-x^2)e^x dx$
1. **Integral 1:**
   \[
   \int \sin x \cos^3 x \, dx
   \]
   We can use the substitution method. Let \( u = \cos x \), then \( du = -\sin x \, dx \). The integral becomes:
   \[
   \int \sin x \cos^3 x \, dx = -\int u^3 \, du
   \]
   Integrating \( u^3 \):
   \[
   -\int u^3 \, du = -\frac{u^4}{4} + C = -\frac{\cos^4 x}{4} + C
   \]
   Therefore, the solution is:
   \[
   \boxed{-\frac{1}{4} \cos^4 x + C}
   \]

2. **Integral 2:**
   \[
   \int \frac{dx}{(1+\sqrt{x})\sqrt{x}}
   \]
   Let \( u = 1 + \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}} \, dx \) or \( 2 \, du = \frac{dx}{\sqrt{x}} \). The integral becomes:
   \[
   2 \int \frac{1}{u} \, du = 2 \ln |u| + C = 2 \ln |1 + \sqrt{x}| + C
   \]
   Therefore, the solution is:
   \[
   \boxed{2 \ln (1 + \sqrt{x}) + C}
   \]

3. **Integral 3:**
   \[
   \int x^2 \sqrt{x^3 + 1} \, dx
   \]
   Let \( u = x^3 + 1 \), then \( du = 3x^2 \, dx \) or \( \frac{1}{3} \, du = x^2 \, dx \). The integral becomes:
   \[
   \frac{1}{3} \int \sqrt{u} \, du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{9} (x^3 + 1)^{3/2} + C
   \]
   Therefore, the solution is:
   \[
   \boxed{\frac{2}{9} (x^3 + 1)^{3/2} + C}
   \]

4. **Integral 4:**
   \[
   \int \frac{e^{2x} - 3e^x}{e^x} \, dx
   \]
   Simplify the integrand:
   \[
   \int (e^x - 3) \, dx
   \]
   Integrate term by term:
   \[
   \int e^x \, dx - \int 3 \, dx = e^x - 3x + C
   \]
   Therefore, the solution is:
   \[
   \boxed{e^x - 3x + C}
   \]

5. **Integral 5:**
   \[
   \int (1 - x^2) e^x \, dx
   \]
   We use integration by parts. Let \( u = 1 - x^2 \) and \( dv = e^x \, dx \). Then \( du = -2x \, dx \) and \( v = e^x \). Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
   \[
   \int (1 - x^2) e^x \, dx = (1 - x^2) e^x - \int e^x (-2x) \, dx
   \]
   Simplify the remaining integral:
   \[
   (1 - x^2) e^x + 2 \int x e^x \, dx
   \]
   Use integration by parts again for \( \int x e^x \, dx \). Let \( u = x \) and \( dv = e^x \, dx \). Then \( du = dx \) and \( v = e^x \):
   \[
   \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C
   \]
   Therefore:
   \[
   \int (1 - x^2) e^x \, dx = (1 - x^2) e^x + 2 (x e^x - e^x) + C = (1 - x^2) e^x + 2x e^x - 2e^x + C
   \]
   Simplify:
   \[
   \int (1 - x^2) e^x \, dx = e^x (1 - x^2 + 2x - 2) + C = e^x (-(x - 1)^2) + C
   \]
   Therefore, the solution is:
   \[
   \boxed{-(x - 1)^2 e^x + C}
   \]