`(5x)/((x+1)(x^(2)-4))`

Explanation

`int(5x)/((x+1)(x^(2)-4))`
`" "=int(5x)/((x+1)(x+2)(x-2))dx`
माना `(5x)/((x+1)(x-2)(x+2))`
`" "=(A)/((x+1))+(B)/((x+2))+(C)/((x-2))" …(1)"`
`" "A(x+2)(x-2)+B(x+1)`
`" "((x-2)+C(x+1)(x+2))/((x+1)(x+2)(x-2))`
`rArr 5x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)`
`{:(x=-1,"तो ",-5=A(1)(-3)+0+0,rArr.,A=(5)/(3)),(x=-2,"तो ",-10=0+B(-1)(-4)+0 ,rArr.,B=-(5)/(2)),(x=2,"तो ",10=0+0+C(3)(4),rArr.,C=(5)/(6)):}`
`therefore" "int(5x)/((x+1)(x-2)(x+2))dx`
`" "=int(5)/(3(x+1))dx-int(5)/(2(x+2))dx+int(5)/(6(x-2))dx`
`=(5)/(3)int(1)/((x+1))dx-(5)/(2)int(1)/((x+2))dx+(5)/(6)int(1)/((x-2))dx`
`=(5)/(3)log|x+1|-(5)/(2)log|x+2|+(5)/(6)log|x-2|+C_(1)`