Question

Resolve into partial fractions of 9/(x-1)(x+3)²​

Answer 1

EXPLANATION.

⇒ ∫9/(x - 1)(x + 3)².

As we know that,

⇒ ∫9/(x - 1)(x + 3)² = ∫A/(x - 1) + ∫B/(x + 3)².

⇒ 9/(x - 1)(x + 3)² = A(x + 3)² + B(x - 1)/(x - 1)(x + 3)².

⇒ 9 = A(x + 3)² + B(x - 1).

As we know that,

Put the value of x = -3 in equation, we get.

⇒ 9 = A(-3 + 3)² + B(-3 - 1).

⇒ 9 = B(-4).

⇒ B = -9/4.

Put the value of x = 1 in equation, we get.

⇒ 9 = A(1 + 3)² + 0.

⇒ 9 = 16A.

⇒ A = 9/16.

Put the value of A & B in equation, we get.

⇒ ∫9/16/(x - 1) + ∫-9/4/(x + 3)².

⇒ ∫9/16(x - 1) + ∫-9/4(x + 3)².

⇒ 9/16∫1/(x - 1) + -9/4∫1/(x + 3)².

⇒ 9/16㏒(x - 1) - 9/4㏒(x + 3)² + c.

                                                                                       

MORE INFORMATION.

Integration of a function.

∫f(x)dx = Ф(x) + c ⇔ d [Ф(x)]/dx = f(x).

Basic theorems on integration.

If f(x), g(x) are two functions of a variable x and k is a constant, then.

(1) = ∫k f(x)dx = k ∫f(x)dx + c.

(2) = ∫ [f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx + c.

(3) = d[∫f(x)dx]/dx = f(x).

(4) = ∫[d f(x)/dx]dx = f(x).

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\tt \ \: :  ⟼ Resolve \: in \: to \: partial \: fractions$

$\tt \ \: :  ⟼ \dfrac{9}{(x - 1) {(x + 3)}^{2} }$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\tt \: Let \dfrac{9}{(x - 1) {( { x+ 3)}}^{2} } = \dfrac{A}{x - 1} + \dfrac{B}{x + 3} + \dfrac{C}{ {(x + 3)}^{2} } - - (1)$

☆ On taking LCM, we get

$\tt \: 9 = A {(x + 3)}^{2} + B(x - 1)(x + 3) +C(x - 1) - - (2)$

❶ On Substituting 'x = 1' in equation (2), we get

$\tt \ \: :  ⟼ 9 = A {(3 + 1)}^{2}$

$\tt \ \: :  ⟼ 9 = 16A$

$\tt \boxed{ \red{ \: :  ⟼ A = \dfrac{9}{16} }}$

❷ Substituting 'x = - 3' in the equation (2), we get

$\tt \ \: :  ⟼ 9 = C( - 3 - 1)$

$\tt \ \: :  ⟼ 9 = - 4C$

$\tt \boxed{ \red {\: :  ⟼ C = - \dfrac{9}{4} }}$

❸ Substituting 'x = 0' in the equation (2), we get

$\tt \ \: :  ⟼ 9 = A {(0 + 3)}^{2} + B(0 - 1)(0 + 3) + C(0 - 1)$

$\tt \ \: :  ⟼ 9 = 9A - 3B - C$

$\tt \ \: :  ⟼ 9 = 9 \times \dfrac{9}{16} - 3B + \dfrac{9}{4}$

$\tt \ \: :  ⟼ 3B = \dfrac{81}{16} + \dfrac{9}{4} - 9$

$\tt \ \: :  ⟼ 3B = \dfrac{81 + 36 - 144}{16}$

$\tt \ \: :  ⟼ 3B = - \dfrac{27}{16}$

$\tt \boxed{ \red{\: :  ⟼ B = - \dfrac{9}{16} }}$

$\tt Hence, \: \dfrac{9}{(x - 1) {(x + 3)}^{2} } = \dfrac{9}{16(x - 1)} - \dfrac{9}{16(x + 3)} - \dfrac{9}{4 {(x + 3)}^{2} }$