Question

\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\frac{x^2 \left( x^4 + 4 \right)}{\left( x^2 + 4 \right)} dx\]
\[ = \int\left( \frac{x^6 + 4 x^2}{x^2 + 4} \right) dx\]
\[\text{ Now }, \]

\[\text{ Therefore }, \frac{x^2 \left( x^4 + 4 \right)}{\left( x^2 + 4 \right)} = \left( x^4 - 4 x^2 + 20 \right) - \frac{80}{x^2 + 4}\]
\[I = \int\frac{x^2 \left( x^4 + 4 \right)}{\left( x^2 + 4 \right)} dx\]
\[ = \int\left( x^4 - 4 x^2 + 20 \right) dx - 80\int\frac{dx}{x^2 + 2^2}\]
\[ = \int x^4 dx - 4\int x^2 dx + 20\int dx - 80\int\frac{dx}{x^2 + 2^2}\]
\[ = \frac{x^{4 + 1}}{4 + 1} - 4 \left[ \frac{x^3}{3} \right] + 20 \left( x \right) - 80 \times \frac{1}{2} \text{ tan }^{- 1} \left( \frac{x}{2} \right) + C\]
\[ = \frac{x^5}{5} - \frac{4}{3} x^3 + 20x - 40 \text{ tan }^{- 1} \left( \frac{x}{2} \right) + C\]