Question

\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text

{dx\]

Answer 1

\[\text{ We have}, \]
\[I = \int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)}\text{ dx }\]
\[ = \int\left( \frac{- x^2 + 1}{- 2 x^2 + x} \right)dx\]
\[ = ∫\frac{1}{2}dx + \int\left( \frac{1 - \frac{x}{2}}{- 2 x^2 + x} \right) d x\]
\[ = \frac{1}{2}\int dx + \frac{1}{2}\int\frac{2 - x}{- 2 x^2 + x}dx\]
\[ = \frac{1}{2}\left[ \int dx + \int\frac{2 - x}{- 2 x^2 + x}dx \right]\]
\[ = \frac{1}{2}\left[ I_1 + I_2 \right] \left( \text{ say }\right)\]
\[\text{ where }I_1 = \int\ dx\ \text{and}\ I_2 = \int\frac{2 - x}{- 2 x^2 + x}dx\]
\[Now, I_1 = \int dx\]
\[ = x + C_1 \]
\[ I_2 = \int\frac{2 - x}{- 2 x^2 + x}dx\]
\[\text{ Let 2 }- x = \text{ A }\frac{d}{dx} \left( - 2 x^2 + x \right) + B\]
\[ \Rightarrow 2 - x = A \left( - 4x + 1 \right) + B\]
\[ \Rightarrow 2 - x = - 4Ax + A + B\]

Comparing coefficients of like terms

\[- 1 = - 4 A\]
\[ \Rightarrow A = \frac{1}{4}\]
\[\text{ and } A + B = 2\]
\[ \Rightarrow \frac{1}{4} + B = 2\]
\[ \Rightarrow B = 2 - \frac{1}{4}\]
\[ = \frac{8 - 1}{4}\]
\[ = \frac{7}{4}\]

\[\therefore \int\frac{2 - x}{- 2 x^2 + x}dx = \int\frac{\frac{1}{4}\left( - 4x + 1 \right) + \frac{7}{4}}{- 2 x^2 + x}dx\]
\[ = \int\frac{\frac{1}{4}\left( - 4x + 1 \right)}{- 2 x^2 + x}dx + \int\frac{\frac{7}{4}}{- 2 x^2 + x}dx\]
\[ = \frac{1}{4}\text{ log } \left| - 2 x^2 + x \right| + \frac{7}{4}\int\frac{1}{- 2\left( x^2 - \frac{x}{2} + \frac{1}{16} - \frac{1}{16} \right)}dx\]
\[ = \frac{1}{4}\text{ log } \left| - 2 x^2 + x \right| - \frac{7}{8}\int\frac{1}{\left\{ \left( x - \frac{1}{4} \right)^2 - \left( \frac{1}{4} \right)^2 \right\}}dx\]
\[ = \frac{1}{4}\text{ log } \left| x\left( - 2x + 1 \right) \right| - \frac{7}{8} \times \frac{1}{2 \times \frac{1}{4}}\text{ log }\left| \frac{x - \frac{1}{4} - \frac{1}{4}}{x - \frac{1}{4} + \frac{1}{4}} \right| + C_2 \]
\[ = \frac{1}{4}\text{ log } \left| x \right| + \frac{1}{4}\text{ log }\left| - 2x + 1 \right| - \frac{7}{4}\log\left| \frac{x - \frac{1}{2}}{x} \right| + C_2 \]
\[ = \frac{1}{4}\text{ log } \left| x \right| + \frac{1}{4}\text{ log } \left| - 2x + 1 \right| - \frac{7}{4}\text{ log } \left| x - \frac{1}{2} \right| + \frac{7}{4}\text{ log } \left| x \right| + C_2 \]
\[ = 2 \text{ log } \left| x \right| + \frac{1}{4}\text{ log } \left| - 2x + 1 \right| - \frac{7}{4}\text{ log } \left| 2x - 1 \right| + C_3 , \text{ where }C_3 = C_2 + \frac{7}{4}\text{ log } 2\]
\[ = 2 \text{ log } \left| x \right| + \frac{1}{4}\text{ log }\left| - 2x + 1 \right| - \frac{7}{4}\text{ log } \left| 1 - 2x \right| + C_3 \]
\[ = 2 \text{ log } \left| x \right| - \frac{3}{2}\text { log } \left| 1 - 2x \right| + C_3 \]
\[\text{ Thus }, I = \frac{1}{2}\left[ x + C_1 + 2 \text{ log } \left| x \right| - \frac{3}{2}\text{ log } \left| 1 - 2x \right| + C_3 \right]\]
\[ = \frac{1}{2}x + \text{ log } \left| x \right| - \frac{3}{4}\text{ log } \left| 1 - 2x \right| + C, \text{ where  C } = \frac{1}{2}\left[ C_1 + C_3 \right]\]