CBSE Class 12 · 2005 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{2 x - 1}{( x - 1 ) ( x + 2 ) ( x - 3 )} d x

Answer: $- \frac{1}{6} lo g ∣ x - 1∣ - \frac{1}{3} lo g ∣ x + 2∣ + \frac{1}{2} lo g ∣ x - 3∣ + C$

Solution

Resolve into partial fractions over the three distinct linear factors:

\frac{2 x - 1}{( x - 1 ) ( x + 2 ) ( x - 3 )} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{D}{x - 3} .

Cover-up at

x = 1, - 2, 3

gives

A = - \frac{1}{6}, B = - \frac{1}{3}, D = \frac{1}{2}

. Then

\int [- \frac{1}{6 ( x - 1 )} - \frac{1}{3 ( x + 2 )} + \frac{1}{2 ( x - 3 )}] d x .

- \frac{1}{6} lo g ∣ x - 1∣ - \frac{1}{3} lo g ∣ x + 2∣ + \frac{1}{2} lo g ∣ x - 3∣ + C

CBSE Class 12 Board · 2005 · Subjective