Mathematics: Will mark if you show work

Step-by-step explanation:Partial fraction:Given integration is (x²-4)= (x+2)(x-2)LetTherefore,A+B=0 ⇒2A+2B=0...(1) and 2B-2A=1.....(2)Adding (1) and (2)2A+2B+2B-2A=0+1⇒4B=1Putting the value of B in equation (1) we getTherefore, [ c is arbitrary constant]Trigonometric substitution:Given integration is Let x = 2 sinθ⇒dx = 2 cosθ dθ [ multiply the numerator and denominator by ]let , [ c is arbitrary constant]Since Putting the value of secθ and tanθ...

Will mark if you show work - id6483991, 902coo

Ответ:

igtguith

27.05.2023

Mathematics

$I=\int \frac{5 \ dx}{x^2-4}$ $=-\frac54 ln |x+2|+\frac54 ln |x+2|+c$

Step-by-step explanation:

Partial fraction:

Given integration is

$I=\int \frac{5 \ dx}{x^2-4}$

(x²-4)= (x+2)(x-2)

$\frac{1}{x^2-4}=\frac{1}{(x+2)(x-2)}$

Let

$\frac{1}{(x+2)(x-2)}=\frac A{x+2}+\frac{B}{x-2}$

$\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ A(x-2)+B(x+2)}{(x+2)(x-2}$

$\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ Ax-2A+Bx+2B}{(x+2)(x-2)}$

$\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ (A+B)x-2A+2B}{(x+2)(x-2)}$

$\Rightarrow 1 = (A+B)x-2A+2B$

Therefore,

A+B=0 ⇒2A+2B=0...(1) and 2B-2A=1.....(2)

Adding (1) and (2)

2A+2B+2B-2A=0+1

⇒4B=1

$\Rightarrow B=\frac14$

Putting the value of B in equation (1) we get

$A=-B=-\frac14$

Therefore,

$I=\int \frac{5 \ dx}{x^2-4}$

$=5[\int -\frac{1}{4(x+2)}dx+\int \frac{1}{4(x-2)}dx]$

$=-\frac54 ln |x+2|+\frac54 ln |x+2|+c$ [ c is arbitrary constant]

Trigonometric substitution:

Given integration is

$I=\int \frac{5 \ dx}{x^2-4}$

Let x = 2 sinθ

⇒dx = 2 cosθ dθ

$= \int \frac{5}{(2sin\theta)^2-4}dx$

$=\int \frac{5}{-4(1-sin^2\theta)}(2 cos \theta) d\theta$

$=-\frac52\int \frac{cos \theta}{cos^2\theta}d\theta$

$=-\frac52\int \frac{1}{cos\theta}d\theta$

$=-\frac52\int sec\theta \ d\theta$

$=-\frac52\int sec\theta\times \frac{(sec\theta+tan\theta)}{(sec\theta+tan\theta)}d\theta$ [ multiply the numerator and denominator by $(sec\theta+tan\theta)$ ]