integration
1/x^2+2x-3×dx
Answers
EXPLANATION.
⇒ ∫dx/x² + 2x - 3.
As we know that,
If coefficient of denominator > coefficient of numerator then we apply partial fractions.
Factorizes the equation into middle term splits, we get.
⇒ x² + 2x - 3.
⇒ x² + 3x - x - 3.
⇒ x(x + 3) - 1(x + 3).
⇒ (x - 1)(x + 3).
⇒ ∫dx/(x - 1)(x + 3).
⇒ 1/(x - 1)(x + 3) = A/(x - 1) + B/(x + 3).
⇒ 1 = A(x + 3) + B(x - 1).
As we know that,
Put the value of x = 1 in equation, we get.
⇒ 1 = A(1 + 3) + B(1 - 1).
⇒ 1 = A(4) + 0.
⇒ 1 = 4A.
⇒ A = 1/4.
Put the value of x = - 3 in equation, we get.
⇒ 1 = A( - 3 + 3) + B(- 3 - 1).
⇒ 1 = 0 + B(-4).
⇒ 1 = - 4B.
⇒ B = -1/4.
Put the values in the equation, we get.
⇒ ∫dx/(x - 1)(x + 3) = ∫A/(x - 1)dx + ∫B/(x + 3)dx.
⇒ ∫dx/(x - 1)(x + 3) = ∫1/4(x - 1)dx + ∫-1/4(x + 3)dx.
⇒ ∫dx/(x - 1)(x + 3) = 1/4∫dx/(x - 1) - 1/4∫dx/(x + 3).
⇒ ∫dx/(x - 1)(x + 3) = 1/4㏑(x - 1) - 1/4㏑(x + 3) + c.
⇒ ∫dx/(x - 1)(x + 3) = 1/4[㏑(x - 1) - ㏑(x + 3)] + c.
MORE INFORMATION.
Integration by parts.
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.
∴ First we arrange the functions in the order according to letters of this word and then integrste by parts.
(2) = If the integrals is of the form,
⇒ ∫eˣ[f(x) + f'(x)]dx = eˣf(x) + c.
(3) = If the integrals is of the form,
⇒ ∫[xf'(x) + f(x)]dx = x f(x) + c.