Evaluate
`int(x^(6)-1)/((x^(2)+1))dx`

Answer

To evaluate the integral

\[ \int \frac{x^6 - 1}{x^2 + 1} \, dx, \]

we will first perform polynomial long division on the integrand.

### Step 1: Polynomial Long Division
We divide \(x^6 - 1\) by \(x^2 + 1\).

1. Divide the leading term: \(x^6 \div x^2 = x^4\).
2. Multiply \(x^4\) by \(x^2 + 1\):
\[ x^4(x^2 + 1) = x^6 + x^4. \]
3. Subtract this from \(x^6 - 1\):
\[ (x^6 - 1) - (x^6 + x^4) = -x^4 - 1. \]

4. Now, divide \(-x^4\) by \(x^2 + 1\):
- Divide the leading term: \(-x^4 \div x^2 = -x^2\).
- Multiply \(-x^2\) by \(x^2 + 1\):
\[ -x^2(x^2 + 1) = -x^4 - x^2. \]
- Subtract:
\[ (-x^4 - 1) - (-x^4 - x^2) = x^2 - 1. \]

5. Now, divide \(x^2 - 1\) by \(x^2 + 1\):
- Divide the leading term: \(x^2 \div x^2 = 1\).
- Multiply \(1\) by \(x^2 + 1\):
\[ 1(x^2 + 1) = x^2 + 1. \]
- Subtract:
\[ (x^2 - 1) - (x^2 + 1) = -2. \]

So, we can express the integral as:

\[ \int \frac{x^6 - 1}{x^2 + 1} \, dx = \int \left( x^4 - x^2 + 1 - \frac{2}{x^2 + 1} \right) \, dx. \]

### Step 2: Integrate Each Term
Now we can integrate term by term:

1. \(\int x^4 \, dx = \frac{x^5}{5}\).
2. \(\int -x^2 \, dx = -\frac{x^3}{3}\).
3. \(\int 1 \, dx = x\).
4. \(\int -\frac{2}{x^2 + 1} \, dx = -2 \tan^{-1}(x)\).

Putting it all together, we have:

\[ \int \frac{x^6 - 1}{x^2 + 1} \, dx = \frac{x^5}{5} - \frac{x^3}{3} + x - 2 \tan^{-1}(x) + C, \]

where \(C\) is the constant of integration.

### Final Answer
Thus, the final result is:

\[ \int \frac{x^6 - 1}{x^2 + 1} \, dx = \frac{x^5}{5} - \frac{x^3}{3} + x - 2 \tan^{-1}(x) + C. \]