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Question
Solution
\[\int\frac{\left( 2x - 3 \right) dx}{x^2 + 6x + 13}\]
\[2x - 3 = A\frac{d}{dx}\left( x^2 + 6x + 13 \right) + B\]
\[2x - 3 = A \left( 2x + 6 \right) + B\]
\[2x - 3 = \left( 2 A \right) x + 6A + B\]
Comparing Coefficients of like powers of x
\[2A = 2\]
\[A = 1\]
\[6 A + B = - 3\]
\[6 + B = - 3\]
\[B = - 9\]
\[ \therefore 2x - 3 = 1 \left( 2x + 6 \right) - 9\]
\[\therefore \int\frac{\left( 2x - 3 \right)}{x^2 + 6x + 13}dx\]
\[ = \int\left( \frac{2x + 6 - 9}{x^2 + 6x + 13} \right)dx\]
` = ∫ ( {2x + 6+ 9}/{x^2 + 6x + 13} ) dx - ∫ {9 dx }/ {x^2 + 6x + 13} `
\[ = \int\frac{\left( 2x + 6 \right) dx}{x^2 + 6x + 13} - 9\int\frac{dx}{x^2 + 6x + 3^2 - 3^2 + 13}\]
\[ = \int\frac{\left( 2x + 6 \right) dx}{x^2 + 6x + 13} - 9\int\frac{dx}{\left( x + 3 \right)^2 + 2^2}\]
\[ = \text{ log } \left| x^2 + 6x + 13 \right| - 9 \times \frac{1}{2} \text{ tan}^{- 1} \left( \frac{x + 3}{2} \right) + C\]
\[ = \text{ log }\left| x^2 + 6x + 13 \right| - \frac{9}{2} \text{ tan}^{- 1} \left( \frac{x + 3}{2} \right) + C\]
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