1) Resolve into partial fraction
2x-1
(x - 4)(3x + 1)
Answers
EXPLANATION.
⇒ ∫(2x - 1)dx/(x - 4)(3x + 1).
As we know that,
Partial fraction is apply only when coefficient of denominator > coefficient of numerator.
Apply partial fraction in this equation, we get.
⇒ (2x - 1)/(x - 4)(3x + 1) = A/(x - 4) + B/(3x + 1).
⇒ (2x - 1) = A(3x + 1) + B(x - 4).
Put the value of x = 4 in equation, we get.
⇒ 2(4) - 1 = A(3(4) + 1) + B(4 - 4).
⇒ 8 - 1 = A(12 + 1) + 0.
⇒ 7 = A(13).
⇒ A = 7/13.
Put the value of x = -1/3 in equation, we get.
⇒ [2(-1/3) - 1] = A[3(-1/3) + 1] + B[-1/3 - 4].
⇒ [-2/3 - 1] = 0 + B[-1 - 12/3].
⇒ [-2 - 3/3] = B[-13/3].
⇒ [-5/3] = B[-13/3].
⇒ -5 = B[-13].
⇒ 5 = 13B.
⇒ B = 5/13.
Put the value in the equation, we get.
⇒ ∫(2x - 1)dx/(x - 4)(3x + 1) = ∫A/(x - 4)dx + ∫B/(3x + 1)dx.
⇒ ∫7/13/(x - 4)dx + ∫5/13/(3x + 1)dx.
⇒ ∫7/13(x - 4)dx + ∫5/13(3x + 1)dx.
⇒ 7/13 ∫dx/(x - 4) + 5/13 ∫dx/(3x + 1).
⇒ 7/13㏑(x - 4) + 5/13㏑(3x + 1) + c.
MORE INFORMATION.
Integration by substitution.
(1) = When integrand is a function of function
that is = ∫f[Ф(x)]Ф'(x)dx.
Here we put Ф(x) = t so that Ф'(x)dx = dt and in that case the integrand is reduced to ∫f(t)dt.
(2) = When integrand is the product of two factors such that one is the derivative of the other that is,
I = ∫f'(x)f(x)dx.
In this case we put f(x) = t and convert it into a standard integrals.
(3) = Integral of a function of the form f(ax + b).
Here we put ax + b = t and convert it into standard integral. obviously if,
∫f(x)dx = Ф(x), then ∫f(ax + b)dx = 1/aФ(ax + b) + c.
EXPLANATION.
⇒ ∫(2x - 1)dx/(x - 4)(3x + 1).
As we know that,
Partial fraction is apply only when coefficient of denominator > coefficient of numerator.
Apply partial fraction in this equation, we get.
⇒ (2x - 1)/(x - 4)(3x + 1) = A/(x - 4) + B/(3x + 1).
⇒ (2x - 1) = A(3x + 1) + B(x - 4).
Put the value of x = 4 in equation, we get.
⇒ 2(4) - 1 = A(3(4) + 1) + B(4 - 4).
⇒ 8 - 1 = A(12 + 1) + 0.
⇒ 7 = A(13).
⇒ A = 7/13.
Put the value of x = -1/3 in equation, we get.
⇒ [2(-1/3) - 1] = A[3(-1/3) + 1] + B[-1/3 - 4].
⇒ [-2/3 - 1] = 0 + B[-1 - 12/3].
⇒ [-2 - 3/3] = B[-13/3].
⇒ [-5/3] = B[-13/3].
⇒ -5 = B[-13].
⇒ 5 = 13B.
⇒ B = 5/13.
Put the value in the equation, we get.
⇒ ∫(2x - 1)dx/(x - 4)(3x + 1) = ∫A/(x - 4)dx + ∫B/(3x + 1)dx.
⇒ ∫7/13/(x - 4)dx + ∫5/13/(3x + 1)dx.
⇒ ∫7/13(x - 4)dx + ∫5/13(3x + 1)dx.
⇒ 7/13 ∫dx/(x - 4) + 5/13 ∫dx/(3x + 1).
⇒ 7/13㏑(x - 4) + 5/13㏑(3x + 1) + c.
MORE INFORMATION.
Integration by substitution.
(1) = When integrand is a function of function
that is = ∫f[Ф(x)]Ф'(x)dx.
Here we put Ф(x) = t so that Ф'(x)dx = dt and in that case the integrand is reduced to ∫f(t)dt.
(2) = When integrand is the product of two factors such that one is the derivative of the other that is,
I = ∫f'(x)f(x)dx.
In this case we put f(x) = t and convert it into a standard integrals.
(3) = Integral of a function of the form f(ax + b).
Here we put ax + b = t and convert it into standard integral. obviously if,
∫f(x)dx = Ф(x), then ∫f(ax + b)dx = 1/aФ(ax + b) + c.
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