InterviewSolution
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InterviewSolution
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फलन `int((x+1)dx)/(3-2x-2x^(2))` का x के सापेक्ष समाकलन कीजिए। |
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Answer» माना `" "I=int((x+1)dx)/(3-2x-2x^(2))" …(1)"` माना `" "(x+1)=A(d)/(dx)(" हर ") +B` `=A(d)/(dx)(3-2x-2x^(2))+B` `=A(0-2-4x)+B` या `" "(x+1)=-4Ax+(b-2A)` दोनों पक्षों में समान घातीय पदों की तुलना करने पर, `1=-4A rArr A=-(1)/(4)` तथा `B-2A = 1 rArr B=2A+1=2(-(1)/(4))+1` `therefore" "B=(1)/(2)` अतः समीकरण (1 ) से, `I=int(A(d)/(dx)("हर")+B)/(3-2x-2x^(2))dx` `=-(1)/(4)int((-2-4x)dx)/(3-2x-2x^(2))+(1)/(2)int(dx)/((3-2x-2x^(2)))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/((3)/(2)-(x^(2)+x))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/(((3)/(2)+(1)/(4))-(x^(2)+x+(1)/(4)))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/(((sqrt7)/(2))^(2)-(x+(1)/(2))^(2))` माना `(sqrt7)/(2)=a` तथा `x+(1)/(2)=t" "therefore" "dx=dt` `therefore" "I=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dt)/(a^(2)-t^(2))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4xx2a)log((a+t)/(a-t))` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4xx2(sqrt7)/(2))log[(sqrt7+2x+1)/(sqrt7-2x-1)]` `=-(1)/(4)log(3-2x-2x^(2))+(1)/(4sqrt7)log((sqrt7+2x+1)/(sqrt7-2x-1))` |
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