Evaluate: `int((x-1)sqrt(x^4+2x^3-x^2+2x+1))/(x^2(x+1))dx`

Answer

To evaluate the integral

\[ I = \int \frac{(x-1) \sqrt{x^4 + 2x^3 - x^2 + 2x + 1}}{x^2(x+1)} \, dx, \]

we will follow a series of steps to simplify and solve the integral.

### Step 1: Simplify the expression under the square root
First, we notice that the expression under the square root can be factored or simplified. We will rewrite the polynomial:

\[ x^4 + 2x^3 - x^2 + 2x + 1. \]

We can try to factor it or complete the square. After some manipulation, we find that:

\[ x^4 + 2x^3 - x^2 + 2x + 1 = (x^2 + x + 1)^2. \]

### Step 2: Substitute the simplified expression
Now we can substitute this back into our integral:

\[ I = \int \frac{(x-1)(x^2 + x + 1)}{x^2(x+1)} \, dx. \]

### Step 3: Expand the numerator
Next, we expand the numerator:

\[ (x-1)(x^2 + x + 1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1. \]

So, we can rewrite the integral as:

\[ I = \int \frac{x^3 - 1}{x^2(x+1)} \, dx. \]

### Step 4: Split the fraction
We can split the fraction:

\[ I = \int \left( \frac{x^3}{x^2(x+1)} - \frac{1}{x^2(x+1)} \right) \, dx = \int \left( \frac{x}{x+1} - \frac{1}{x^2(x+1)} \right) \, dx. \]

### Step 5: Simplify the first term
The first term can be simplified further:

\[ \frac{x}{x+1} = 1 - \frac{1}{x+1}. \]

Thus, the integral becomes:

\[ I = \int \left( 1 - \frac{1}{x+1} - \frac{1}{x^2(x+1)} \right) \, dx. \]

### Step 6: Integrate term by term
Now we can integrate each term separately:

1. \(\int 1 \, dx = x\)
2. \(\int -\frac{1}{x+1} \, dx = -\ln|x+1|\)
3. For \(\int -\frac{1}{x^2(x+1)} \, dx\), we can use partial fractions:

\[ -\frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}. \]

Solving for \(A\), \(B\), and \(C\) gives us:

\[ -1 = A x(x+1) + B(x+1) + Cx^2. \]

Setting \(x = 0\) gives \(B = -1\). Setting \(x = -1\) gives \(A = 0\). Setting \(x = 1\) gives \(C = 1\). Thus:

\[ -\frac{1}{x^2(x+1)} = -\frac{1}{x^2} + \frac{1}{x+1}. \]

Therefore, we have:

\[ I = x - \ln|x+1| + \int \left(-\frac{1}{x^2} + \frac{1}{x+1}\right) \, dx. \]

### Step 7: Final integration
Now we can integrate:

1. \(\int -\frac{1}{x^2} \, dx = \frac{1}{x}\)
2. \(\int \frac{1}{x+1} \, dx = \ln|x+1|\)

Putting it all together:

\[ I = x - \ln|x+1| - \frac{1}{x} + \ln|x+1| + C. \]

The \(-\ln|x+1|\) and \(+\ln|x+1|\) cancel out:

\[ I = x - \frac{1}{x} + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ \int \frac{(x-1) \sqrt{x^4 + 2x^3 - x^2 + 2x + 1}}{x^2(x+1)} \, dx = x - \frac{1}{x} + C. \]