1/x^+3x+2 resolve into partial fraction
Answers
EXPLANATION.
⇒ ∫dx/x² + 3x + 2.
As we know that,
First we factorizes the denominator, we get.
Factorizes the equation into middle term split, we get.
⇒ x² + 3x + 2.
⇒ x² + 2x + x + 2.
⇒ x(x + 2) + 1(x + 2).
⇒ (x + 1)(x + 2).
Put this value in equation, we get.
⇒ ∫dx/(x + 1)(x + 2). = A/(x + 1) + B/(x + 2).
Taking L.C.M on both sides, we get.
⇒ 1 = A(x + 2) + B(x + 1).
Put the value of x = -2 in equation, we get.
⇒ 1 = A(-2 + 2) + B(-2 + 1).
⇒ 1 = 0 + B(-1).
⇒ B = -1.
Put the value of x = -1 in equation, we get.
⇒ 1 = A(-1 + 2) + B(-1 + 1).
⇒ 1 = A(1) + 0.
⇒ A = 1.
Put the value of A = 1 & B = -1 in equation, we get.
⇒ ∫A/(x + 1) + ∫B/(x + 2).
⇒ ∫1.dx/(x + 1) + ∫(-1).dx/(x + 2).
⇒ ㏑(x + 1) - ㏑(x + 2) + c.
MORE INFORMATION.
Standard integrals.
(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) + c = aˣ㏒(e) + c.
Answer :-
⇒ ∫dx/x² + 3x + 2.
As we know that,
First we factorizes the denominator, we get.
Factorizes the equation into middle term split, we get.
⇒ x² + 3x + 2.
⇒ x² + 2x + x + 2.
⇒ x(x + 2) + 1(x + 2).
⇒ (x + 1)(x + 2).
Put this value in equation, we get.
⇒ ∫dx/(x + 1)(x + 2). = A/(x + 1) + B/(x + 2).
Taking L.C.M on both sides, we get.
⇒ 1 = A(x + 2) + B(x + 1).
Put the value of x = -2 in equation, we get.
⇒ 1 = A(-2 + 2) + B(-2 + 1).
⇒ 1 = 0 + B(-1).
⇒ B = -1.
Put the value of x = -1 in equation, we get.
⇒ 1 = A(-1 + 2) + B(-1 + 1).
⇒ 1 = A(1) + 0.
⇒ A = 1.
Put the value of A = 1 & B = -1 in equation, we get.
⇒ ∫A/(x + 1) + ∫B/(x + 2).
⇒ ∫1.dx/(x + 1) + ∫(-1).dx/(x + 2).
⇒ ㏑(x + 1) - ㏑(x + 2) + c.