Question

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

Answer 1

 

\[\int\frac{\text{ 2x } dx}{\left( 2 + x - x^2 \right)}\]
\[2x = A\frac{d}{dx}\left( 2 + x - x^2 \right) + B\]
\[2x = A \left( 0 + 1 - 2x \right) + B\]
\[2x = \left( - 2 A \right) x + A + B\]

Comparing the Coefficients of like powers of x

\[- 2\text{ A }= 2\]
\[A = - 1\]
\[A + B = 0\]
\[ - 1 + B = 0\]
\[B = 1\]

Now ` ∫  { 2x    dx }/ {(2 + x - x^2 )}`

`=∫   ({-1 ( 1 - 2x ) + 1 } / { -x^2 + x + 2 }) dx`
\[ = - \int\left( \frac{1 - 2x}{- x^2 + x + 2} \right)dx + \int\frac{dx}{- x^2 + x + 2}\]
\[ = - I_1 + I_2 . . . \left( 1 \right) \left( say \right) where\]
\[ I_1 = \int\left( \frac{1 - 2x}{- x^2 + x + 2} \right)dx\]
\[ I_2 = \int\frac{dx}{- x^2 + x + 2}\]
\[ I_1 = \int\left( \frac{1 - 2x}{- x^2 + x + 2} \right)dx\]
\[\text{ let }- x^2 + x + 2 = t\]
\[ \Rightarrow \left( 1 - 2x \right) dx = dt\]
\[ I_1 = \int\frac{dt}{t}\]
\[ I_1 = \text{ log } \left| t \right| + C_1 \]
\[ = \text{ log } \left| 2 + x - x^2 \right| + C_1 . . . \left( 2 \right)\]
\[ I_2 = \int\frac{dx}{- x^2 + x + 2}\]
\[ I_2 = \int\frac{- dx}{x^2 - x - 2}\]
\[ I_2 = \int\frac{- dx}{x^2 - x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 - 2}\]
\[ I_2 = \int\frac{- dx}{\left( x - \frac{1}{2} \right)^2 - \left( \frac{3}{2} \right)^2}\]
\[ I_2 = - \frac{1}{2 \times \frac{3}{2}}\text{ log }\left| \frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}} \right| + C_2 \]
\[ I_2 = - \frac{1}{3} \text{ log }\left| \frac{x - 2}{x + 1} \right| + C_2 . . . \left( 3 \right)\]
\[\text{ from } \left( 1 \right) \left( 2 \right)\text{ and }\left( 3 \right)\]
\[\int\left( \frac{2x}{2 + x - x^2} \right)dx = - \text{ log } \left| 2 + x - x^2 \right| - \frac{1}{3}\text{ log }\left| \frac{x - 2}{x + 1} \right| + C_1 + C_2 \]
\[ = - \text{ log } \left| 2 + x - x^2 \right| + \frac{1}{3} \log \left| \frac{1 + x}{x - 2} \right| + C\]
\[\text{ where } C = C_1 + C_2\]