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

Answer

To evaluate the integral

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

we will follow a systematic approach.

### Step 1: Rewrite the integrand

We can rewrite the integrand by adding and subtracting 1 in the numerator:

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

### Step 2: Simplify the first term

Next, we focus on simplifying the first term:

\[ \frac{x^6 + 1}{x^2 + 1}. \]

Notice that \(x^6 + 1\) can be factored as:

\[ x^6 + 1 = (x^2)^3 + 1^3 = (x^2 + 1)(x^4 - x^2 + 1). \]

So, we can write:

\[ \frac{x^6 + 1}{x^2 + 1} = x^4 - x^2 + 1. \]

### Step 3: Substitute back into the integral

Now substituting this back into our integral, we have:

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

### Step 4: Separate the integrals

We can now separate the integral into simpler parts:

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

### Step 5: Evaluate each integral

Now, we will evaluate each integral separately.

1. For \(\int x^4 \, dx\):

\[ \int x^4 \, dx = \frac{x^5}{5}. \]

2. For \(\int x^2 \, dx\):

\[ \int x^2 \, dx = \frac{x^3}{3}. \]

3. For \(\int 1 \, dx\):

\[ \int 1 \, dx = x. \]

4. For \(\int \frac{1}{x^2 + 1} \, dx\):

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

### Step 6: Combine the results

Now, substituting these results back into the expression for \(I\):

\[ I = \left( \frac{x^5}{5} - \frac{x^3}{3} + x \right) - 2 \tan^{-1}(x) + C, \]

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

### Final Answer

Thus, the final result for the integral is:

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

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