`int log (5+t) dt`

Answer

To solve the integral \( \int \log(5+t) \, dt \), we will follow these steps:

### Step 1: Substitution
Let \( x = 5 + t \). Then, we differentiate both sides to find \( dt \):
\[ dx = dt \implies dt = dx \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral in terms of \( x \):
\[ \int \log(5+t) \, dt = \int \log(x) \, dx \]

### Step 3: Integration by Parts
We will use integration by parts to solve \( \int \log(x) \, dx \). Recall the formula for integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Here, we choose:
- \( u = \log(x) \) (thus \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (thus \( v = x \))

### Step 4: Apply Integration by Parts
Now we apply the integration by parts formula:
\[ \int \log(x) \, dx = x \log(x) - \int x \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ \int \log(x) \, dx = x \log(x) - \int 1 \, dx \]
\[ = x \log(x) - x + C \]

### Step 5: Substitute Back
Now we substitute back \( x = 5 + t \):
\[ \int \log(5+t) \, dt = (5+t) \log(5+t) - (5+t) + C \]
\[ = (5+t) \log(5+t) - (5+t) + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \log(5+t) \, dt = (5+t) \log(5+t) - (5+t) + C \]