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Question
\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
Sum
Solution
\[\int \left( \frac{x^6 + 1}{x^2 + 1} \right)dx\]
\[ = \int \left[ \frac{\left( x^2 \right)^3 + 1^3}{x^2 + 1} \right]\text{dx }A^3 + B^3 = \left( A + B \right) \left( A^2 - AB + B^2 \right)\]
\[ = \int\frac{\left( x^2 + 1 \right)\left( x^4 - x^2 + 1 \right)}{\left( x^2 + 1 \right)}dx\]
\[ = \int\left( x^4 - x^2 + 1 \right)dx\]
\[ = \int x^4 dx + \int x^2 dx + \int1dx\]
\[ = \frac{x^{4 + 1}}{4 + 1} - \frac{x^{2 + 1}}{2 + 1} + x + C\]
\[ = \frac{x^5}{5} - \frac{x^3}{3} + x + C\]
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