CBSE Class 12 · 2010 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{1 - x ^{2}}{x ( 1 - 2 x )} d x

Answer: $\frac{1}{2} x + lo g ∣ x ∣ - \frac{3}{4} lo g ∣1 - 2 x ∣ + C$

Solution

The fraction is improper, so divide. Writing it over

2 x^{2} - x

\frac{1 - x ^{2}}{x ( 1 - 2 x )} = \frac{x ^{2} - 1}{2 x ^{2} - x} = \frac{1}{2} (1 + \frac{x - 2}{2 x ^{2} - x}) .

Partial fractions on

\frac{x - 2}{x ( 2 x - 1 )}

give

\frac{2}{x} + \frac{3}{1 - 2 x}

, so

\frac{1}{2} \int (1 + \frac{2}{x} + \frac{3}{1 - 2 x}) d x = \frac{1}{2} x + lo g ∣ x ∣ - \frac{3}{4} lo g ∣1 - 2 x ∣ + C .

\frac{1}{2} x + lo g ∣ x ∣ - \frac{3}{4} lo g ∣1 - 2 x ∣ + C

CBSE Class 12 Board · 2010 · Subjective