Question

\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\frac{x^2 dx}{x^2 + 6x + 12}\]
\[\text{ Now }, \]

\[\text{ Therefore }, \]
\[\frac{x^2}{x^2 + 6x + 12} = 1 - \frac{\left( 6x + 12 \right)}{x^2 + 6x + 12} . . . . . \left( 1 \right)\]
\[\text { Let 6x } + 12 = A\frac{d}{dx} \left( x^2 + 6x + 12 \right) + B\]
\[ \Rightarrow 6x + 12 = A \left( 2x + 6 \right) + B\]
\[ \Rightarrow 6x + 12 = \left( 2A \right) x + 6A + B\]
\[\text{ Equating Coefficients of like terms }]\]
\[2A = 6\]
\[A = 3\]
\[6A + B = 12\]
\[18 + B = 12\]
\[B = - 6\]
\[ \therefore \frac{x^2}{x^2 + 6x + 12} = 1 - \frac{3 \left( 2x + 6 \right) - 6}{x^2 + 6x + 12}\]
\[I = \int\frac{x^2 dx}{x^2 + 6x + 12}\]
\[ = \int dx - 3\int\frac{\left( 2x + 6 \right) dx}{x^2 + 6x + 12} + 6\int\frac{dx}{x^2 + 6x + 12}\]
\[ = \int dx - 3 \int\frac{\left( 2x + 6 \right) dx}{x^2 + 6x + 12} + 6\int\frac{dx}{x^2 + 6x + 9 + 3}\]
\[ = \int dx - 3\int\frac{\left( 2x + 6 \right) dx}{x^2 + 6x + 12} + 6\int\frac{dx}{\left( x + 3 \right)^2 + \left( \sqrt{3} \right)^2}\]
\[ = x - 3 \text{ log } \left| x^2 + 6x + 12 \right| + \frac{6}{\sqrt{3}} \text{ tan }^{- 1} \left( \frac{x + 3}{\sqrt{3}} \right) + C\]
\[ = x - 3 \text{ log } \left| x^2 + 6x + 12 \right| + 2\sqrt{3} \text{ tan }^{- 1} \left( \frac{x + 3}{\sqrt{3}} \right) + C\]