If `I_(n)= int_(1)^(e ) log^(n) x dx and I_(n)= a +b I_(n-1)` then a= …. and b= …..

Answer

To solve the problem, we need to evaluate the integral \( I_n = \int_1^e \log^n x \, dx \) and express it in the form \( I_n = a + b I_{n-1} \). Let's go through the steps systematically.

### Step 1: Define the integral
We start with the integral:
\[ I_n = \int_1^e \log^n x \, dx \]

### Step 2: Integration by parts
We will use integration by parts, where we let:
- \( u = \log^n x \) (the first function)
- \( dv = dx \) (the second function)

Now, we need to differentiate \( u \) and integrate \( dv \):
- \( du = n \log^{n-1} x \cdot \frac{1}{x} \, dx \)
- \( v = x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[ I_n = \left[ x \log^n x \right]_1^e - \int_1^e x \cdot n \log^{n-1} x \cdot \frac{1}{x} \, dx \]

### Step 3: Evaluate the boundary term
Now we evaluate the boundary term:
\[ \left[ x \log^n x \right]_1^e = \left( e \cdot \log^n e \right) - \left( 1 \cdot \log^n 1 \right) = e \cdot 1^n - 1 \cdot 0 = e - 0 = e \]

### Step 4: Simplify the integral
Now we simplify the integral:
\[ I_n = e - n \int_1^e \log^{n-1} x \, dx \]
This integral is \( I_{n-1} \):
\[ I_n = e - n I_{n-1} \]

### Step 5: Rearranging the equation
Rearranging the equation gives us:
\[ I_n = e - n I_{n-1} \]

### Step 6: Identify constants a and b
From the equation \( I_n = a + b I_{n-1} \), we can identify:
- \( a = e \)
- \( b = -n \)

### Final Result
Thus, the values of \( a \) and \( b \) are:
\[ a = e, \quad b = -n \]