Question

resolve into partial fraction. x+5÷(2x-1) (x+4)​

Answer 1

Answer:

Answer

Let

(x+3)(x+1)

2

(2x−5)

=

x+3

A

+

x+1

B

+

(x+1)

2

C

⇒2x−5=A(x+1)

2

+B(x+1)(x+3)+C(x+3)

⇒2x−5=A(x

2

+2x+1)+B(x

2

+4x+3)+C(x+3)

On comparing coefficients

0=A+B,2=2A+4B+C,−5=A+3B+3C

⇒A=−

4

11

,B=

4

11

,C=−

2

7

Hence,

(x+3)(x+1)

2

(2x−5)

=−

4(x+3)

11

+

4(x+1)

11

−

2(x+1)

2

7

Hence, option 'B' is correct.

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Answer 2

EXPLANATION.

⇒ ∫x + 5/(2x - 1)(x + 4).dx.

As we know that,

Partial fraction is apply only when coefficient of Denominator > coefficient of Numerator.

⇒ ∫x + 5/(2x - 1)(x + 4).dx = A/(2x - 1) + B/(x + 4).

Taking L.C.M on both sides, we get.

⇒ x + 5/(2x - 1)(x + 4) = A/(2x - 1) + B/(x + 4).

⇒ x + 5 = A(x + 4) + B(2x - 1).

Put the value of x = -4 in equation, we get.

⇒ - 4 + 5 = A(- 4 + 4) + B[2(-4) - 1].

⇒ 1 = 0 + B[- 8 - 1].

⇒ 1 = - 9B.

⇒ B = -1/9.

Put the value of x = 1/2 in equation, we get.

⇒ 1/2 + 5 = A(1/2 + 4) + B[2(1/2) - 1].

⇒ 1 + 10/2 = A(1 + 8/2) + 0.

⇒ 11/2 = 9A/2.

⇒ 11 = 9A.

⇒ A = 11/9.

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

⇒ ∫x + 5/(2x - 1)(x + 4).dx = ∫A/(2x - 1).dx + ∫B/(x + 4).dx.

⇒ ∫x + 5/(2x - 1)(x + 4).dx = ∫11/9/(2x - 1).dx + ∫-1/9/(x + 4).dx.

⇒ ∫11/9(2x - 1).dx + ∫-1/9(x + 4).dx.

⇒ 11/9㏑(2x - 1) + (-1)/9㏑(x + 4) + c.

⇒ 11/9㏑(2x - 1) - 1/9㏑(x + 4) + c.

                                                                                                                     

MORE INFORMATION.

Standard integral.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, (k ∈ R).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ -1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = aˣ/㏒ₐ(e) + c. = ㏒ₐ(e) + c.