(1) 2009 Kansai University entrance exam

Calculate $ \int \frac{e^{\minus{}2x}}{1\plus{}e^{\minus{}x}}\ dx$.


(2) 2009 Rikkyo University entrance exam/Science

Evaluate $ \int_0^ 1 \frac{2x^3}{1\plus{}x^2}\ dx$.
### Problem 1: Calculate \( \int \frac{e^{-2x}}{1+e^{-x}} \, dx \)

1. **Substitution**: Let \( u = 1 + e^{-x} \). Then, \( du = -e^{-x} \, dx \) or \( dx = -\frac{du}{e^{-x}} \). Notice that \( e^{-x} = u - 1 \).

2. **Rewrite the integral**:
   \[
   \int \frac{e^{-2x}}{1+e^{-x}} \, dx = \int \frac{e^{-2x}}{u} \cdot \left( -\frac{du}{e^{-x}} \right) = -\int \frac{e^{-2x}}{u} \cdot \frac{du}{e^{-x}}
   \]

3. **Simplify the integrand**:
   \[
   -\int \frac{e^{-2x}}{u} \cdot \frac{du}{e^{-x}} = -\int \frac{e^{-2x}}{u} \cdot \frac{1}{e^{-x}} \, du = -\int \frac{e^{-2x}}{u} \cdot e^x \, du = -\int \frac{e^{-x}}{u} \, du
   \]

4. **Express \( e^{-x} \) in terms of \( u \)**:
   \[
   e^{-x} = u - 1
   \]

5. **Substitute back**:
   \[
   -\int \frac{u-1}{u} \, du = -\int \left( 1 - \frac{1}{u} \right) \, du = -\int 1 \, du + \int \frac{1}{u} \, du
   \]

6. **Integrate**:
   \[
   -u + \ln|u| + C
   \]

7. **Substitute \( u = 1 + e^{-x} \) back**:
   \[
   -(1 + e^{-x}) + \ln|1 + e^{-x}| + C = -1 - e^{-x} + \ln(1 + e^{-x}) + C
   \]

Thus, the integral is:
\[
\boxed{-1 - e^{-x} + \ln(1 + e^{-x}) + C}
\]

### Problem 2: Evaluate \( \int_0^1 \frac{2x^3}{1+x^2} \, dx \)

1. **Rewrite the integrand**:
   \[
   \frac{2x^3}{1+x^2} = 2x \cdot \frac{x^2}{1+x^2}
   \]

2. **Substitute \( u = x^2 \)**: Then, \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \).

3. **Rewrite the integral**:
   \[
   \int_0^1 \frac{2x^3}{1+x^2} \, dx = \int_0^1 2x \cdot \frac{x^2}{1+x^2} \, dx = \int_0^1 2x \cdot \frac{u}{1+u} \cdot \frac{du}{2x} = \int_0^1 \frac{u}{1+u} \, du
   \]

4. **Simplify the integrand**:
   \[
   \int_0^1 \frac{u}{1+u} \, du = \int_0^1 \left( 1 - \frac{1}{1+u} \right) \, du = \int_0^1 1 \, du - \int_0^1 \frac{1}{1+u} \, du
   \]

5. **Integrate**:
   \[
   \int_0^1 1 \, du - \int_0^1 \frac{1}{1+u} \, du = \left[ u \right]_0^1 - \left[ \ln|1+u| \right]_0^1
   \]

6. **Evaluate the definite integrals**:
   \[
   \left[ u \right]_0^1 = 1 - 0 = 1
   \]
   \[
   \left[ \ln|1+u| \right]_0^1 = \ln(1+1) - \ln(1+0) = \ln 2 - \ln 1 = \ln 2 - 0 = \ln 2
   \]

7. **Combine the results**:
   \[
   1 - \ln 2
   \]

Thus, the integral is:
\[
\boxed{1 - \ln 2}
\]