`intx sqrt(x+x^(2))dx` का मान ज्ञात कीजिए ।

Explanation

`intx sqrt(x+x^(2))dx`
यदि `x=A.(d)/(dx)(x+x^(2))+B`
`rArr" "x=A(1+2x)+B" …(1)"`
अब समीकरण (1 ) के दोनों पक्षों में x की घातों की तुलना करने पर
`2A=1 rArr A=(1)/(2)`
व`" "A+B=0 rArr B=-(1)/(2)`
`rArr" "int xsqrt(x+x^(2))dx=int{(1)/(2)(1+2x)-(1)/(2)}sqrt(x+x^(2))dx`
`=(1)/(2)int(1+2x)sqrt(x+x^(2))dx-(1)/(2)intsqrt(x+x^(2))dx`
माना `(x+x^(2))= t rArr (1+2x)dx=dt`, तब
`=(1)/(2)int sqrtt dt-(1)/(2)int sqrt({(x^(2)+x+(1)/(4))-(1)/(4)})dx`
`=(1)/(2).(t^(3//2))/((3//2))-(1)/(2)sqrt({(x+(1)/(2))^(2)-((1)/(2))^(2)})dx`
माना `(x+(1)/(2))=u rArr dx =du,`
तब , `=(1)/(3)(x+x^(2))^(3//2)-(1)/(2).intsqrt(u^(2)-((1)/(2))^(2))du`
`=(1)/(3)(x+x^(2))^(3//2)-(1)/(2){(u)/(2)sqrt(u^(2)-(1)/(4))-(1)/(8)log[u+sqrt(u^(2)-(1)/(4))]}+c`
`[" यहाँ "intsqrt(x^(2)-a^(2))dx=(x)/(2)sqrt(x^(2)-a^(2))-(a^(2))/(2)log[x+sqrt(x^(2)-a^(2))]+c]`
`=(1)/(3)(x+x^(2))^(3//2)-(1)/(4)(x+(1)/(2))sqrt((x+(1)/(2))^(2)-(1)/(4))+(1)/(16)log.[(x+(1)/(2))sqrt((x+(1)/(2))^(2)-(1)/(4))]+c`
`=(1)/(3)(x+x^(2))^(3//2)-(1)/(8)(2x+1)sqrt(x+x^(2))+(1)/(16)log[(2x+1)/(2).sqrt(x+x^(2))]_c`