InterviewSolution
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InterviewSolution
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फलन `int(3x+1)/(2x^(2)+x-1)dx` का x के सापेक्ष समाकलन कीजिए। |
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Answer» माना `" "I=int(3x+1)/(2x^(2)+x-1)dx" …(1)"` माना अंश `=A(d)/(dx)("हर")+B` `rArr 3x+1=A(d)/(dx)(2x^(2)+x-1)+B` `=A(4x+1)+B` ltBrgt `3x+1=4Ax+(A+B)` दोनों पक्षों से समान घातीय पदों के गुणांकों की तुलना करने पर, `3=4A rArr A=(3)/(4)` तथा `A+B=1 rArr B=1-A=1-(3)/(4)=(1)/(4)` अतः समीकरण (1 ) से , `I=int(A(d)/(dx)(" हर")+B)/(2x^(2)+x-1)dx` `=(3)/(4)int((4x+1)dx)/(2x^(2)+x-1)+(1)/(4)int(dx)/(2x^(2)+x-1)` `=(3)/(4)int((4x+1)dx)/(2x^(2)+x-1)dx+(1)/(8)int(dx)/(x^(2)+(1)/(2)x-(1)/(2))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dx)/(x^(2)+(1)/(2)x_(1)/(16)-(1)/(2)-(1)/(16))` `I=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dx)/((x+(1)/(4))^(2)-((3)/(4))^(2))` माना `x+(1)/(4)=t" "therefore" "dx=dt" तथा "(3)/(4)=a` `therefore" "I=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dt)/(t^(2)-a^(2))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(16a)log((t-a)/(t+1))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(16xx(3)/(4))log((x+(1)/(4)-(3)/(4))/(x+(1)/(4)+(3)/(4)))` `=(3)/(4)log(2x^(2)+x-1)+(1)/(12)log((2x-1)/(2(x+1)))` |
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