समाकलन कीजिए!
(x)/((x^(2)+1)(x-1))`

Explanation

`int(x)/((x^(2)+1)(x-1))dx`
माना `(x)/((x^(2)+1)(x-1))=(A)/(x-1)+(Bx+C)/(x^(2)+1)" …(1)"`
`rArr" "x=A(x^(2)+1)+(Bx+C)(x-1)" …(2)"`
`{:(x=1,"तो ",1=A(2)+0,rArr.,A=(1)/(2)),(x=0,"तो ",0=A-C,rArr.,C=A=(1)/(2)):}`
`x^(2)` के गुणांक को बरारबर रखने पर ,
`0=A+B" "rArr" "B=-A=-(1)/(2)`
`therefore" "int(x)/((x^(2)+1)(x-1))dx`
`=int{((1)/(2))/(x-1)+((-(1)/(2)x+(1)/(2)))/(x^(2)+1)}dx`
`=(1)/(2)int(1)/(x-1)dx+(1)/(2)int((-x+1))/(x^(2)+1)dx`
`=(1)/(2)int(1)/(x-1)dx-(1)/(2)int(x)/(x^(2)+1)dx+(1)/(2)int(1)/(x^(2)+1)dx`
`=(1)/(2)log|x-1|-(1)/(4)I_(1)+(1)/(2)tan^(-1)x+C_(1)`
जहाँ, `I_(1)=int(2x)/(x^(2)+1)dx" माना "(x^(2)+1)=t`
`" "rArr" "2xdx=dt`
`I_(1)=int(dt)/(t)=log|t|+C_(2)+log|x^(2)+1|+C_(2)`
समीकरण (3 ) से,
`int(x)/((x^(2)+1)(x-1))dx=(1)/(2)log|x-1|-(1)/(4)log|x^(2)+1|+(1)/(2)tan^(-1)x+C`