Evaluate: $\int \frac{1 - x^{2}}{x (1 - 2 x)} dx$

Solution

$\int \frac{1 - x^{2}}{x (1 - 2 x)} dx Here \frac{1 - x^{2}}{x (1 - 2 x)} is an improper rational fraction .$

Reducing it to proper rational fraction gives

$\frac{1 - x^{2}}{x (1 - 2 x)} = \frac{1}{2} + \frac{1}{2} (\frac{2 - x}{x (1 - 2 x)}) . . . . . . . (i) Now, let \frac{2 - x}{x (1 - 2 x)} = \frac{A}{x} + \frac{B}{(1 - 2 x)} \Rightarrow \frac{2 - x}{x (1 - 2 x)} = \frac{A (1 - 2 x) + Bx}{x (1 - 2 x)} \Rightarrow 2 - x = A - x (2 A - B)$

Equating the coefficients we get, A = 2 and B = 3

$So, \frac{2 - x}{x (1 - 2 x)} = \frac{2}{x} + \frac{3}{(1 - 2 x)}$

Substituting in equation (i), we get

$\frac{1 - x^{2}}{x (1 - 2 x)} = \frac{1}{2} + \frac{1}{2} (\frac{2}{x} + \frac{3}{(1 - 2 x)}) i . e . \int \frac{1 - x^{2}}{x (1 - 2 x)} dx = \int [\frac{1}{2} + \frac{1}{2} (\frac{2}{x} + \frac{3}{(1 - 2 x)})] dx = \int \frac{dx}{2} + \int \frac{dx}{x} + \frac{3}{2} \int \frac{dx}{(1 - 2 x)} = \frac{x}{2} + \log |x| + \frac{3}{2} x \frac{1}{(- 2)} \log |1 - 2 x| + C = \frac{x}{2} + \log |x| - \frac{3}{4} x \log |1 - 2 x| + C .$

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