`(x+2)/(sqrt(x^(2)+2x+3))`

Explanation

माना `I=int(x+2)/(sqrt(x^(2)+2x+3))dx`
माना `x+2=A(d)/(dx)(x^(2)+2x+3)+B`
`rArr" "x+2=A(2x+2)+B`
x के गुणकों की तुलना करने पर,
`1=2A" "rArr" "A=(1)/(2),`
अचर पदों की तुलना करने पर,
`" "2=2A+B`
`rArr" "2=1+B" "rArr" "B=1`
`therefore I=int((1)/(2)(2x+2)+1)/(sqrt(x^(2)+2x+3))dx`
`=(1)/(2)int(2x+2)/(sqrt(x^(2)+2x+3))dx+int(1)/(sqrt(x^(2)+2x+3))dx" "` प्रथम संकलन के लिये
`" माना "x^(2)+2x+3=t`
`" "rArr" "(2x+2)dx=dt`
`=(1)/(2)int(1)/(sqrtt)dt+int(1)/(sqrt((x^(2)+2x+1)+2))dx`
: `=(1)/(2)int t^(-1//2)dt+int(1)/(sqrt((x+1)^(2)+(sqrt2)^(2)))dx`
`=(1)/(2).(t^(1//2))/(1//2)+log|x+1+sqrt((x+1)^(2)+(sqrt2)^(2))|+c`
`=sqrt(x^(2)+2x+3)+log|x+1+sqrt(x^(2)+2x+3)|+c`