integral 6/x²-3x-2 please help to calculate this and explain step by step..
Answers
EXPLANATION.
⇒ ∫6.dx/(x² - 3x - 2).
As we know that,
If the coefficient of Denominator > coefficient of numerator then we can apply partial fractions, we get.
In this question we apply partial fractions, we get.
⇒ x² - 3x - 2.
Factorizes the equation into middle term splits, we get.
⇒ x² - 2x - x - 2.
⇒ x(x - 2) + 1(x - 2).
⇒ (x + 1)(x - 2).
⇒ ∫6.dx/(x + 1)(x - 2).
⇒ ∫6.dx/(x + 1)(x - 2) = ∫A/(x + 1)dx + ∫B/(x - 2)dx.
⇒ 6/(x + 1)(x - 2) = A(x - 2) + B(x + 1)/(x - 2)(x + 1).
⇒ 6 = A(x - 2) + B(x + 1).
As we know that,
Put the value of x = 2 in equation, we get.
⇒ 6 = A(2 - 2) + B(2 + 1).
⇒ 6 = 0 + 3B.
⇒ B = 6/3.
⇒ B = 2.
Put the value of x = - 1 in equation, we get.
⇒ 6 = A(- 1 - 2) + B(- 1 + 1).
⇒ 6 = A(-3) + 0.
⇒ 6 = - 3A.
⇒ A = - 2.
Put the value in the main equation, we get.
⇒ ∫-2.dx/(x + 1) + ∫2.dx/(x - 2).
⇒ -2 ∫dx/(x + 1) + 2 ∫dx/(x - 2).
⇒ - 2 ㏑(x + 1) + 2 ㏑(x - 2) + C.
⇒ 2 ㏑(x - 2) - 2 ㏑(x + 1) + C.
MORE INFORMATION.
Integration by parts.
(1) = If u and v are two functions of x then,
∫(u v)dx = u ∫(v dx) - ∫[(du/dx). (∫v dx)]dx.
From the first letter of the word.
I = Inverse trigonometric functions.
L = Logarithmic functions.
A = Algebraic functions.
T = Trigonometric functions.
E = Exponential functions.
We get a word = ILATE.
Therefore, first arrange the functions in the order according to letters of this word and then integrate by-parts.
(2) = ∫eˣ[f(x) + f'(x)]dx = eˣ f(x) + C.
(3) = ∫[x f'(x) + f(x)]dx = x f(x) + C.