Find the following integrals :
`int (x^3-x^2+x-1)/(x-1) dx`

Answer

To solve the integral \( I = \int \frac{x^3 - x^2 + x - 1}{x - 1} \, dx \), we will first simplify the integrand.

### Step 1: Simplifying the integrand
We can perform polynomial long division on \( x^3 - x^2 + x - 1 \) by \( x - 1 \).

1. Divide \( x^3 \) by \( x \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x - 1 \) to get \( x^3 - x^2 \).
3. Subtract \( (x^3 - x^2) \) from \( (x^3 - x^2 + x - 1) \):
\[ (x^3 - x^2 + x - 1) - (x^3 - x^2) = x - 1 \]
4. Now divide \( x - 1 \) by \( x - 1 \) to get \( 1 \).
5. Multiply \( 1 \) by \( x - 1 \) to get \( x - 1 \).
6. Subtract \( (x - 1) \) from \( (x - 1) \):
\[ (x - 1) - (x - 1) = 0 \]

Thus, we have:
\[ \frac{x^3 - x^2 + x - 1}{x - 1} = x^2 + 1 \]

### Step 2: Setting up the integral
Now we can rewrite the integral:
\[ I = \int (x^2 + 1) \, dx \]

### Step 3: Integrating
Now we can integrate term by term:
\[ I = \int x^2 \, dx + \int 1 \, dx \]

Using the power rule for integration:
1. \(\int x^2 \, dx = \frac{x^3}{3} + C_1\)
2. \(\int 1 \, dx = x + C_2\)

Combining these results:
\[ I = \frac{x^3}{3} + x + C \]
where \( C = C_1 + C_2 \) is the constant of integration.

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{x^3 - x^2 + x - 1}{x - 1} \, dx = \frac{x^3}{3} + x + C \]

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