To solve the integral \(\int \frac{x^6}{x^2 + 1} \, dx\), we can follow these steps:
### Step 1: Rewrite the integrand
We can rewrite \(x^6\) as \(x^6 = x^4 \cdot x^2\), which allows us to express the integrand in a more manageable form:
\[
\int \frac{x^6}{x^2 + 1} \, dx = \int \frac{x^4 \cdot x^2}{x^2 + 1} \, dx
\]
### Step 2: Use polynomial long division
Since the degree of the numerator is greater than the degree of the denominator, we can perform polynomial long division. Dividing \(x^6\) by \(x^2 + 1\) gives:
\[
x^6 \div (x^2 + 1) = x^4 - x^2 + 1 - \frac{1}{x^2 + 1}
\]
Thus, we can rewrite the integral:
\[
\int \frac{x^6}{x^2 + 1} \, dx = \int \left( x^4 - x^2 + 1 - \frac{1}{x^2 + 1} \right) \, dx
\]
### Step 3: Split the integral
Now we can split the integral into separate parts:
\[
\int \left( x^4 - x^2 + 1 - \frac{1}{x^2 + 1} \right) \, dx = \int x^4 \, dx - \int x^2 \, dx + \int 1 \, dx - \int \frac{1}{x^2 + 1} \, dx
\]
### Step 4: Integrate each term
Now we can integrate each term separately:
1. \(\int x^4 \, dx = \frac{x^5}{5}\)
2. \(\int x^2 \, dx = \frac{x^3}{3}\)
3. \(\int 1 \, dx = x\)
4. \(\int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x)\)
### Step 5: Combine the results
Putting it all together, we have:
\[
\int \frac{x^6}{x^2 + 1} \, dx = \frac{x^5}{5} - \frac{x^3}{3} + x - \tan^{-1}(x) + C
\]
### Final Answer
Thus, the final answer is:
\[
\int \frac{x^6}{x^2 + 1} \, dx = \frac{x^5}{5} - \frac{x^3}{3} + x - \tan^{-1}(x) + C
\]
