Answer :
Answer:
[tex]I=\int \frac{5 \ dx}{x^2-4}[/tex] [tex]=-\frac54 ln |x+2|+\frac54 ln |x+2|+c[/tex]
Step-by-step explanation:
Partial fraction:
Given integration is
[tex]I=\int \frac{5 \ dx}{x^2-4}[/tex]
(x²-4)= (x+2)(x-2)
[tex]\frac{1}{x^2-4}=\frac{1}{(x+2)(x-2)}[/tex]
Let
[tex]\frac{1}{(x+2)(x-2)}=\frac A{x+2}+\frac{B}{x-2}[/tex]
[tex]\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ A(x-2)+B(x+2)}{(x+2)(x-2}[/tex]
[tex]\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ Ax-2A+Bx+2B}{(x+2)(x-2)}[/tex]
[tex]\Rightarrow \frac{1}{(x+2)(x-2)}=\frac{ (A+B)x-2A+2B}{(x+2)(x-2)}[/tex]
[tex]\Rightarrow 1 = (A+B)x-2A+2B[/tex]
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
[tex]\Rightarrow B=\frac14[/tex]
Putting the value of B in equation (1) we get
[tex]A=-B=-\frac14[/tex]
Therefore,
[tex]I=\int \frac{5 \ dx}{x^2-4}[/tex]
[tex]=5[\int -\frac{1}{4(x+2)}dx+\int \frac{1}{4(x-2)}dx][/tex]
[tex]=-\frac54 ln |x+2|+\frac54 ln |x+2|+c[/tex] [ c is arbitrary constant]
Trigonometric substitution:
Given integration is
[tex]I=\int \frac{5 \ dx}{x^2-4}[/tex]
Let x = 2 sinθ
⇒dx = 2 cosθ dθ
[tex]= \int \frac{5}{(2sin\theta)^2-4}dx[/tex]
[tex]=\int \frac{5}{-4(1-sin^2\theta)}(2 cos \theta) d\theta[/tex]
[tex]=-\frac52\int \frac{cos \theta}{cos^2\theta}d\theta[/tex]
[tex]=-\frac52\int \frac{1}{cos\theta}d\theta[/tex]
[tex]=-\frac52\int sec\theta \ d\theta[/tex]
[tex]=-\frac52\int sec\theta\times \frac{(sec\theta+tan\theta)}{(sec\theta+tan\theta)}d\theta[/tex] [ multiply the numerator and denominator by [tex](sec\theta+tan\theta)[/tex] ]
[tex]=-\frac52\int \frac{(sec^2\theta+sec\theta tan\theta)}{(sec\theta+tan\theta)}d\theta[/tex]
let [tex]f(x)= sec\theta+tan \theta[/tex] ,[tex]f'(\theta)= (sec^2\theta+sec\theta tan\theta)d\theta[/tex]
[tex]=-\frac52\int \frac{f'(\theta)}{f(\theta)} d\theta[/tex]
[tex]=-\frac52 ln (f(\theta))+c[/tex] [ c is arbitrary constant]
[tex]=-\frac52 ln(sec\theta+tan\theta)+c[/tex]
Since [tex]x= 2 sin \theta[/tex]
[tex]\Rightarrow sin \theta =\frac{x}{2}[/tex]
[tex]\Rightarrow cos \theta = \sqrt{1-(\frac x2)^2[/tex]
[tex]\Rightarrow cos \theta =\frac{\sqrt{4-x^2}}{2}[/tex]
[tex]\therefore tan \theta =\frac{sin\theta}{cos\theta}= \frac{x}{\sqrt{4-x^2}}[/tex]
[tex]\therefore sec\theta = \frac{2}{\sqrt{4-x^2}}[/tex]
Putting the value of secθ and tanθ
[tex]=-\frac52 ln| \frac{2}{\sqrt{4-x^2}}+\frac{x}{\sqrt{4-x^2}}|+c[/tex]
[tex]=-\frac52 ln |\frac{x+2}{\sqrt{4-x^2}}|+c[/tex]
[tex]=-\frac52 ln |x+2|+\frac54 ln |4-x^2|+c[/tex]
[tex]=-\frac54 ln |x+2|+\frac54 ln |x+2|+c[/tex]