Question

\[\int\frac{x + 2}{2 x^2 + 6x + 5}\text{  dx }\]

Answer 1

\[\int\left( \frac{x + 2}{2 x^2 + 6x + 5} \right)dx\]
\[x + 2 = A\frac{d}{dx}\left( 2 x^2 + 6x + 5 \right) + B\]
\[x + 2 = A \left( 4x + 6 \right) + B\]
\[x + 2 = \left( 4 A \right) x + 6 A + B\]

Comparing the Coefficients of like powers of x

\[\text{ 4 A }= 1\]
\[A = \frac{1}{4}\]
\[\text{ 6 A + B } = 2\]
\[6 \times \frac{1}{4} + B = 2\]
\[B = \frac{1}{2}\]

\[\therefore \int\left( \frac{x + 2}{2 x^2 + 6x + 5} \right)dx\]
\[ = \int\left[ \frac{\frac{1}{4}\left( 4x + 6 \right) + \frac{1}{2}}{2 x^2 + 6x + 5} \right]dx\]
\[ = \frac{1}{4}\int\frac{\left( 4x + 6 \right)}{2 x^2 + 6x + 5}dx + \frac{1}{2}\int\frac{1}{2 x^2 + 6x + 5}dx\]
\[ = \frac{1}{4}\int\frac{\left( 4x + 6 \right)}{2 x^2 + 6x + 5}dx + \frac{1}{4}\int\frac{dx}{x^2 + 3x + \frac{5}{2}}\]
\[ = \frac{1}{4}\int\frac{\left( 4x + 6 \right)}{2 x^2 + 6x + 5}dx + \frac{1}{4}\int\frac{dx}{x^2 + 3x + \left( \frac{3}{2} \right)^2 - \left( \frac{3}{2} \right)^2 + \frac{5}{2}}\]
\[ = \frac{1}{4}\int\frac{\left( 4x + 6 \right)}{2 x^2 + 6x + 5}dx + \frac{1}{4}\int\frac{dx}{\left( x + \frac{3}{2} \right)^2 - \frac{9}{4} + \frac{5}{2}}\]
\[ = \frac{1}{4}\int\frac{\left( 4x + 6 \right)}{2 x^2 + 6x + 5}dx + \frac{1}{4}\int\frac{dx}{\left( x + \frac{3}{2} \right)^2 + \frac{1}{4}}\]
\[ = \frac{1}{4}\int\frac{\left( 4x + 6 \right)}{2 x^2 + 6x + 5}dx + \frac{1}{4}\int\frac{dx}{\left( x + \frac{3}{2} \right)^2 + \left( \frac{1}{2} \right)^2}\]
\[ = \frac{1}{4} \text{  log }\left| 2 x^2 + 6x + 5 \right| + \frac{1}{4} \times 2 \text{ tan}^{- 1} \left( \frac{x + \frac{3}{2}}{\frac{1}{2}} \right) + C\]
\[ = \frac{1}{4} \text{ log }\left| 2 x^2 + 6x + 5 \right| + \frac{1}{2} \text{ tan}^{- 1} \left( 2x + 3 \right) + C\]