x^2+1/(x-2)^2(1-2x) resolve into partial fraction
Answers
EXPLANATION.
⇒ ∫x² + 1/(x - 2)²(1 - 2x)dx.
As we know that,
Partial fractions is applied only when coefficient of denominator > coefficient of numerator.
It is a type of repeated linear factors.
⇒ ∫x² + 1/(x - 2)²(1 - 2x)dx. = A/(x - 2) + B/(x - 2)² + C/(1 - 2x).
⇒ x² + 1 = A(x - 2)(1 - 2x) + B(1 - 2x) + C(x - 2)².
As we know that,
Put the value of x = 2 in equation, we get.
⇒ (2)² + 1 = A(2 - 2)(1 - 2(2)) + B(1 - 2(2)) + C(2 - 2)².
⇒ 4 + 1 = 0 + B(1 - 4) + 0.
⇒ 5 = -3B.
⇒ B = -5/3.
Put the value of x = 1/2 in equation, we get.
⇒ (1/2)² + 1 = A[1/2 - 2][1 - 2(1/2)] + B[1 - 2(1/2)] + C[1/2 - 2]².
⇒ 1/4 + 1 = 0 + 0 + C[1 - 4/2]².
⇒ 1 + 4/4 = C[-3/2]².
⇒ 5/4 = C[9/4].
⇒ 5 = 9C.
⇒ C = 5/9.
Put the value of x = 0 in equation, we get.
⇒ (0)² + 1 = A(0 - 2)(1 - 2(0)) + B(1 - 2(0)) + C(0 - 2)².
⇒ 1 = A[-2][1] + B[1] + C[-2]².
⇒ 1 = -2A + B + 4C.
⇒ 1 = -2A + [-5/3] + 4[5/9].
⇒ 1 = -2A - 5/3 + 20/9.
⇒ 1 = -18A - 15 + 20/9.
⇒ 9 = -18A + 5.
⇒ 9 - 5 = -18A.
⇒ 4 = -18A.
⇒ 2 = -9A.
⇒ A = -2/9.
Put the value in equation, we get.
⇒ ∫x² + 1/(x - 2)²(1 - 2x)dx = ∫A/(x - 2)dx + ∫B/(x - 2)²dx + ∫C/(1 - 2x)dx.
⇒ ∫-2/9/(x - 2)dx + ∫-5/3/(x - 2)²dx + ∫5/9/(1 - 2x)dx.
⇒ ∫-2/9(x - 2)dx + ∫-5/3(x - 2)²dx + ∫5/9(1 - 2x)dx.
⇒ -2/9 ∫dx/(x - 2) - 5/3 ∫dx/(x - 2)² + 5/9 ∫dx/(1 - 2x).
⇒ -2/9㏑(x - 2) - 5/3 [x - 2]⁻¹/1 + 5/9㏑(1 - 2x) + C.
MORE INFORMATION.
Standard form of integrals.
(1) = Standard forms of integrals.
(a) = ∫f'(x)/f(x)dx = ㏒[f(x)] + c.
(b) = [f(x)]ⁿf'(x)dx = [f(x)]ⁿ⁺¹/n + 1 + c (provided n ≠ -1).
(c) = ∫f'(x)/√f(x)dx = 2√f(x) + c.
(2) = Integrals in the form = ∫dx/a sin(x) + b cos(x).
Putting a = r cos∅ and b = r sin∅, we get.
I = ∫dx/r sin(x + ∅) = 1/r ∫cosec(x + ∅)dx. = 1/r㏒ tan(x/2 + ∅/2) + c = 1/√a² + b² ㏒ tan(x/2 + 1/2tan⁻¹b/a) + c.
EXPLANATION.
⇒ ∫x² + 1/(x - 2)²(1 - 2x)dx.
As we know that,
Partial fractions is applied only when coefficient of denominator > coefficient of numerator.
It is a type of repeated linear factors.
⇒ ∫x² + 1/(x - 2)²(1 - 2x)dx. = A/(x - 2) + B/(x - 2)² + C/(1 - 2x).
⇒ x² + 1 = A(x - 2)(1 - 2x) + B(1 - 2x) + C(x - 2)².
As we know that,
Put the value of x = 2 in equation, we get.
⇒ (2)² + 1 = A(2 - 2)(1 - 2(2)) + B(1 - 2(2)) + C(2 - 2)².
⇒ 4 + 1 = 0 + B(1 - 4) + 0.
⇒ 5 = -3B.
⇒ B = -5/3.
Put the value of x = 1/2 in equation, we get.
⇒ (1/2)² + 1 = A[1/2 - 2][1 - 2(1/2)] + B[1 - 2(1/2)] + C[1/2 - 2]².
⇒ 1/4 + 1 = 0 + 0 + C[1 - 4/2]².
⇒ 1 + 4/4 = C[-3/2]².
⇒ 5/4 = C[9/4].
⇒ 5 = 9C.
⇒ C = 5/9.
Put the value of x = 0 in equation, we get.
⇒ (0)² + 1 = A(0 - 2)(1 - 2(0)) + B(1 - 2(0)) + C(0 - 2)².
⇒ 1 = A[-2][1] + B[1] + C[-2]².
⇒ 1 = -2A + B + 4C.
⇒ 1 = -2A + [-5/3] + 4[5/9].
⇒ 1 = -2A - 5/3 + 20/9.
⇒ 1 = -18A - 15 + 20/9.
⇒ 9 = -18A + 5.
⇒ 9 - 5 = -18A.
⇒ 4 = -18A.
⇒ 2 = -9A.
⇒ A = -2/9.
Put the value in equation, we get.
⇒ ∫x² + 1/(x - 2)²(1 - 2x)dx = ∫A/(x - 2)dx + ∫B/(x - 2)²dx + ∫C/(1 - 2x)dx.
⇒ ∫-2/9/(x - 2)dx + ∫-5/3/(x - 2)²dx + ∫5/9/(1 - 2x)dx.
⇒ ∫-2/9(x - 2)dx + ∫-5/3(x - 2)²dx + ∫5/9(1 - 2x)dx.
⇒ -2/9 ∫dx/(x - 2) - 5/3 ∫dx/(x - 2)² + 5/9 ∫dx/(1 - 2x).
⇒ -2/9㏑(x - 2) - 5/3 [x - 2]⁻¹/1 + 5/9㏑(1 - 2x) + C.
MORE INFORMATION.
Standard form of integrals.
(1) = Standard forms of integrals.
(a) = ∫f'(x)/f(x)dx = ㏒[f(x)] + c.
(b) = [f(x)]ⁿf'(x)dx = [f(x)]ⁿ⁺¹/n + 1 + c (provided n ≠ -1).
(c) = ∫f'(x)/√f(x)dx = 2√f(x) + c.
(2) = Integrals in the form = ∫dx/a sin(x) + b cos(x).
Putting a = r cos∅ and b = r sin∅, we get.
I = ∫dx/r sin(x + ∅) = 1/r ∫cosec(x + ∅)dx. = 1/r㏒ tan(x/2 + ∅/2) + c = 1/√a² + b² ㏒ tan(x/2 + 1/2tan⁻¹b/a) + c.