Question

resolve the partial fraction​

Answer 1

Given :

$\sf \dfrac{x}{(x + 1)(x - 2)(x - 1)}$

$\sf \dfrac{x}{(x + 1)(x - 2)(x - 1)} = \frac{A}{(x + 1)} + \frac{B}{(x - 2)} + \frac{C}{(x - 1)}$

• To find the value of 'A', put x = -1 except in (x+1)
• To find the of 'B', put the value of x = 2 except in (x-2)
• To find the value of 'C', put x = 1 except in (x-1)

$\therefore \: A = \frac{ - 1}{6} \: \: B = \frac{2}{3} \: \: C = \frac{ - 1}{2}$

$\frac{x}{(x + 1)(x - 2)(x - 1)} = \frac{ - 1}{6(x + 1)} + \frac{2}{3(x - 2)} - \frac{1}{2(x - 1)}$

$\int \frac{x}{(x + 1)(x - 2)(x - 1)} = \int\left( \frac{ - 1}{6(x + 1)} + \frac{2}{3(x - 2)} - \frac{1}{2(x - 1)}\right)dx$

$\int \frac{x}{(x + 1)(x - 2)(x - 1)} = - \frac{1}{6} \int\frac{ 1}{(x + 1)} dx + \frac{2}{3} \int \frac{1}{(x - 2)} dx - \frac{1}{2} \int\frac{1}{(x - 1)}dx$

$\sf\int \dfrac{x}{(x + 1)(x - 2)(x - 1)} = - \frac{1}{6}log|x + 1 |+ \frac{2}{3} log|x - 2|- \frac{1}{2} log|x - 1|$

Points to remember when we do partial function integration :

1. P(x)/(x+a)(x+b) = A/(x+a) + B/(x+b)
2. P(x)/(x²+a)(x+b) = Ax+B/(x²+a) + C/(x+b)
3. P(x)/(x+a)²(x+b) = A/(x+a)+B(x+a)² + C(x+b)

Answer 2

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