Show that : `int_0^1(logx)/((1+x))dx=-int_0^1(log(1+x))/x dx`

Explanation

To show that

\[ \int_0^1 \frac{\log x}{1+x} \, dx = -\int_0^1 \frac{\log(1+x)}{x} \, dx, \]

we will use integration by parts. Let's proceed step by step.

### Step 1: Define the integral
Let

\[ I = \int_0^1 \frac{\log x}{1+x} \, dx. \]

### Step 2: Apply integration by parts
We will use integration by parts, which states:

\[ \int u \, dv = uv - \int v \, du. \]

Here, we can choose:

- \( u = \log x \) (thus \( du = \frac{1}{x} \, dx \))
- \( dv = \frac{1}{1+x} \, dx \) (thus \( v = \log(1+x) \))

### Step 3: Compute \( uv \) and the integral
Now we compute \( uv \):

\[ uv = \log x \cdot \log(1+x) \Big|_0^1. \]

Evaluating this at the limits:

- At \( x = 1 \): \( \log(1) \cdot \log(2) = 0 \cdot \log(2) = 0 \).
- At \( x = 0 \): \( \log(0) \cdot \log(1) \) approaches \( -\infty \cdot 0 \), which is an indeterminate form. However, we can analyze the limit more carefully.

Using L'Hôpital's Rule or recognizing that \( \log x \to -\infty \) and \( \log(1+x) \to 0 \) as \( x \to 0 \), we find that this term approaches \( 0 \).

Thus,

\[ uv \Big|_0^1 = 0 - 0 = 0. \]

### Step 4: Compute the remaining integral
Now we need to compute the remaining integral:

\[ -\int_0^1 \log(1+x) \cdot \frac{1}{x} \, dx. \]

So we have:

\[ I = 0 - \int_0^1 \frac{\log(1+x)}{x} \, dx = -\int_0^1 \frac{\log(1+x)}{x} \, dx. \]

### Conclusion
Thus, we have shown that:

\[ \int_0^1 \frac{\log x}{1+x} \, dx = -\int_0^1 \frac{\log(1+x)}{x} \, dx. \]