Evaluate `int_(5)^(7)ln(x-3)^(2)dx+2int_(0)^(1)ln(x+4)^(2)dx`

Answer

To evaluate the integral \( \int_{5}^{7} \ln(x-3)^{2} \, dx + 2 \int_{0}^{1} \ln(x+4)^{2} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrals using properties of logarithms
Using the property \( \ln(a^b) = b \ln(a) \), we can rewrite the integrals:
\[ \int_{5}^{7} \ln(x-3)^{2} \, dx = 2 \int_{5}^{7} \ln(x-3) \, dx \]
\[ 2 \int_{0}^{1} \ln(x+4)^{2} \, dx = 4 \int_{0}^{1} \ln(x+4) \, dx \]
Thus, the expression becomes:
\[ 2 \int_{5}^{7} \ln(x-3) \, dx + 4 \int_{0}^{1} \ln(x+4) \, dx \]

### Step 2: Evaluate the first integral \( \int \ln(x-3) \, dx \)
Using integration by parts, let:
- \( u = \ln(x-3) \) then \( du = \frac{1}{x-3} \, dx \)
- \( dv = dx \) then \( v = x \)

Applying integration by parts:
\[ \int \ln(x-3) \, dx = x \ln(x-3) - \int \frac{x}{x-3} \, dx \]
Now, simplify \( \int \frac{x}{x-3} \, dx \):
\[ \int \frac{x}{x-3} \, dx = \int \left( 1 + \frac{3}{x-3} \right) \, dx = x + 3 \ln(x-3) \]
Thus, we have:
\[ \int \ln(x-3) \, dx = x \ln(x-3) - (x + 3 \ln(x-3)) = (x - 3) \ln(x-3) - x \]

### Step 3: Evaluate the definite integral from 5 to 7
Now we compute:
\[ \int_{5}^{7} \ln(x-3) \, dx = \left[ (x-3) \ln(x-3) - x \right]_{5}^{7} \]
Calculating at the limits:
- At \( x = 7 \):
\[ (7-3) \ln(7-3) - 7 = 4 \ln(4) - 7 \]
- At \( x = 5 \):
\[ (5-3) \ln(5-3) - 5 = 2 \ln(2) - 5 \]
Thus:
\[ \int_{5}^{7} \ln(x-3) \, dx = \left[ 4 \ln(4) - 7 \right] - \left[ 2 \ln(2) - 5 \right] \]
Simplifying this gives:
\[ = 4 \ln(4) - 2 \ln(2) - 2 \]
Using \( \ln(4) = 2 \ln(2) \):
\[ = 4(2 \ln(2)) - 2 \ln(2) - 2 = 8 \ln(2) - 2 \ln(2) - 2 = 6 \ln(2) - 2 \]

### Step 4: Evaluate the second integral \( \int \ln(x+4) \, dx \)
Using integration by parts again:
Let:
- \( u = \ln(x+4) \) then \( du = \frac{1}{x+4} \, dx \)
- \( dv = dx \) then \( v = x \)

Thus:
\[ \int \ln(x+4) \, dx = x \ln(x+4) - \int \frac{x}{x+4} \, dx \]
Simplifying \( \int \frac{x}{x+4} \, dx \):
\[ \int \frac{x}{x+4} \, dx = \int \left( 1 - \frac{4}{x+4} \right) \, dx = x - 4 \ln(x+4) \]
So:
\[ \int \ln(x+4) \, dx = x \ln(x+4) - (x - 4 \ln(x+4)) = 5 \ln(x+4) - x \]

### Step 5: Evaluate the definite integral from 0 to 1
Now compute:
\[ \int_{0}^{1} \ln(x+4) \, dx = \left[ 5 \ln(x+4) - x \right]_{0}^{1} \]
Calculating at the limits:
- At \( x = 1 \):
\[ 5 \ln(1+4) - 1 = 5 \ln(5) - 1 \]
- At \( x = 0 \):
\[ 5 \ln(0+4) - 0 = 5 \ln(4) \]
Thus:
\[ \int_{0}^{1} \ln(x+4) \, dx = (5 \ln(5) - 1) - 5 \ln(4) = 5 \ln(5) - 5 \ln(4) - 1 \]
Using \( \ln(4) = 2 \ln(2) \):
\[ = 5(\ln(5) - 2 \ln(2)) - 1 \]

### Step 6: Combine results
Now substitute back into the original expression:
\[ 2 \int_{5}^{7} \ln(x-3) \, dx + 4 \int_{0}^{1} \ln(x+4) \, dx \]
This becomes:
\[ 2(6 \ln(2) - 2) + 4(5(\ln(5) - 2 \ln(2)) - 1) \]
Calculating:
\[ = 12 \ln(2) - 4 + 20 \ln(5) - 8 \ln(2) - 4 = 4 \ln(2) + 20 \ln(5) - 8 \]

### Final Answer
Thus, the final result is:
\[ 4 \ln(2) + 20 \ln(5) - 8 \]