Value of `int (x ^(2) + 1)/((x -1) (x-2)) dx` is

Options

A. `x + log [ ((x-2)^(5))/( (x -1) ^(2))] +C`
B. `x + log [ (( x -1) ^(2))/( (x -2 ) ^(5))] +C`
C. `x - log [ ((x -1) ^(5))/( (x -1) ^(1))] + C`
D. None of these

Answer

To solve the integral \( \int \frac{x^2 + 1}{(x - 1)(x - 2)} \, dx \), we will follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand:
\[ \frac{x^2 + 1}{(x - 1)(x - 2)} = \frac{x^2 - 1 + 2}{(x - 1)(x - 2)} = \frac{(x^2 - 1) + 2}{(x - 1)(x - 2)} \]
Using the difference of squares, we can express \( x^2 - 1 \) as \( (x - 1)(x + 1) \):
\[ = \frac{(x - 1)(x + 1) + 2}{(x - 1)(x - 2)} \]

### Step 2: Split the fraction
Now we can split the fraction into two parts:
\[ = \frac{(x - 1)(x + 1)}{(x - 1)(x - 2)} + \frac{2}{(x - 1)(x - 2)} \]
This simplifies to:
\[ = \frac{x + 1}{x - 2} + \frac{2}{(x - 1)(x - 2)} \]

### Step 3: Integrate each term separately
Now we can integrate each term separately:
\[ \int \left( \frac{x + 1}{x - 2} + \frac{2}{(x - 1)(x - 2)} \right) \, dx = \int \frac{x + 1}{x - 2} \, dx + \int \frac{2}{(x - 1)(x - 2)} \, dx \]

### Step 4: Solve the first integral
For the first integral \( \int \frac{x + 1}{x - 2} \, dx \), we can use substitution:
Let \( u = x - 2 \) then \( du = dx \) and \( x = u + 2 \):
\[ \int \frac{(u + 2) + 1}{u} \, du = \int \left( 1 + \frac{3}{u} \right) \, du = u + 3 \ln |u| + C_1 = (x - 2) + 3 \ln |x - 2| + C_1 \]

### Step 5: Solve the second integral using partial fractions
For the second integral \( \int \frac{2}{(x - 1)(x - 2)} \, dx \), we use partial fractions:
\[ \frac{2}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \]
Multiplying through by the denominator gives:
\[ 2 = A(x - 2) + B(x - 1) \]
Setting \( x = 1 \):
\[ 2 = A(1 - 2) + B(1 - 1) \Rightarrow 2 = -A \Rightarrow A = -2 \]
Setting \( x = 2 \):
\[ 2 = A(2 - 2) + B(2 - 1) \Rightarrow 2 = B \Rightarrow B = 2 \]
Thus:
\[ \frac{2}{(x - 1)(x - 2)} = \frac{-2}{x - 1} + \frac{2}{x - 2} \]
Now we can integrate:
\[ \int \left( \frac{-2}{x - 1} + \frac{2}{x - 2} \right) \, dx = -2 \ln |x - 1| + 2 \ln |x - 2| + C_2 \]

### Step 6: Combine results
Combining both integrals, we have:
\[ \int \frac{x^2 + 1}{(x - 1)(x - 2)} \, dx = (x - 2) + 3 \ln |x - 2| - 2 \ln |x - 1| + 2 \ln |x - 2| + C \]
This simplifies to:
\[ = x - 2 + 5 \ln |x - 2| - 2 \ln |x - 1| + C \]

### Final Answer
The value of the integral is:
\[ \int \frac{x^2 + 1}{(x - 1)(x - 2)} \, dx = x - 2 + 5 \ln |x - 2| - 2 \ln |x - 1| + C \]