If `int_(0)^(1) (e^(x))/( 1+x) dx = k`, then `int_(0)^(1) (e^(x))/( (1+x)^(2)) dx` is equal to

Options

A. `k-1+(e )/(2)`
B. `k + 1 - ( e )/( 2)`
C. `k - 1 - ( e)/(2)`
D. `k+1+(e )/(2)`

Answer

To solve the problem, we need to evaluate the integral

\[ \int_{0}^{1} \frac{e^x}{(1+x)^2} \, dx \]

given that

\[ \int_{0}^{1} \frac{e^x}{1+x} \, dx = k. \]

### Step-by-Step Solution:

1. **Identify the Given Integral**:
We know that
\[ \int_{0}^{1} \frac{e^x}{1+x} \, dx = k. \]

2. **Use Integration by Parts**:
We can use integration by parts on the integral we want to evaluate. We can set:
- \( u = \frac{1}{1+x} \) (which is our first preference, an algebraic function)
- \( dv = e^x \, dx \) (which is our second preference, an exponential function)

Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = -\frac{1}{(1+x)^2} \, dx \]
- Integrate \( dv \):
\[ v = e^x \]

3. **Apply Integration by Parts Formula**:
The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du. \]
Substituting our values:
\[ \int \frac{e^x}{1+x} \, dx = \left[ \frac{e^x}{1+x} \right]_{0}^{1} - \int e^x \left(-\frac{1}{(1+x)^2}\right) \, dx. \]

4. **Evaluate the Boundary Term**:
Now we evaluate the boundary term:
\[ \left[ \frac{e^x}{1+x} \right]_{0}^{1} = \frac{e^1}{1+1} - \frac{e^0}{1+0} = \frac{e}{2} - 1. \]

5. **Substitute Back into the Integral**:
Thus, we have:
\[ \int_{0}^{1} \frac{e^x}{1+x} \, dx = \left( \frac{e}{2} - 1 \right) + \int_{0}^{1} \frac{e^x}{(1+x)^2} \, dx. \]
Setting this equal to \( k \):
\[ k = \frac{e}{2} - 1 + \int_{0}^{1} \frac{e^x}{(1+x)^2} \, dx. \]

6. **Rearranging to Solve for the Desired Integral**:
Rearranging gives us:
\[ \int_{0}^{1} \frac{e^x}{(1+x)^2} \, dx = k - \left( \frac{e}{2} - 1 \right). \]
Simplifying further:
\[ \int_{0}^{1} \frac{e^x}{(1+x)^2} \, dx = k + 1 - \frac{e}{2}. \]

### Final Answer:
Thus, the value of the integral

\[ \int_{0}^{1} \frac{e^x}{(1+x)^2} \, dx = k - \frac{e}{2} + 1. \]