InterviewSolution
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InterviewSolution
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`int(x+3)sqrt(3-4x-x^(2)).dx` का मान ज्ञात कीजिए । |
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Answer» यदि `" "I=int(x+3)srt(3-4x-x^(2))dx` अब `" "(d)/(dx)(3-4x-x^(2))=-4 - 2x` अतः `" "x+3 =-(1)/(2)(-4 -2x)+1` `therefore I=intsqrt(3-4x-x^(2)).{-(1)/(2)(-4-2x)+1}dx` `=-(1)/(2)intsqrt(3-4x-x^(2))(-4 - 2x)dx+intsqrt(3-4x-x^(2))dx` `=-(1)/(2)I_(1)+I_(2)` यहाँ `I_(1)=intsqrt(3-4x-x^(2)).(-4-2x)dx` `3-4x-x^(2)=t rArr -4-2xdx=dt` `therefore" "I_(1)=intt^(1//2)dt=(t^(3//2))/(3//2)=(2)/(3)(3-4x-x^(2))^(3//2)" ...(2)"` तथा `" "I_(2)=intsqrt(3-4x-x^(2))dx` `=intsqrt(7-(x+2)^(2))dx` `=intsqrt((sqrt7)^(2)-(x+2)^(2))dx` `=((x+2)sqrt(3-4x-x^(2)))/(2)+(7)/(2)sin^(-1)((x+2)/(sqrt7))+c" ...(3)"` अब, समीकरण (1 ), (2 ) व (3 ) से `I=-(1)/(3)(3-4x-x^(2))^(3//2)+(1)/(2)(x+2)sqrt(3-4x-x^(2))+(7)/(2)sin^(-1)((x+2)/(sqrt7))+c` |
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