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प्रश्न
`int (2x - 7)/sqrt(4x- 1) dx`
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उत्तर
Let `I = int (2x - 7)/sqrt(4x - 1) dx`
Let 2x − 7 = A(4x − 1) + B
∴ 2x – 7 = 4Ax + (B – A)
By equating the coefficients on both sides, we get
4A = 2 and B – A = –7
∴ A = `1/2` and B = –7 + A
= `-7 + 1/2`
= `-13/2`
∴ `2x - 7 = 1/2(4x - 1) - 13/2`
∴ `I = 1/2 int ((4x - 1) - 13)/sqrt(4x - 1) dx`
= `1/2 int((4x - 1)/sqrt(4x - 1) - 13/sqrt(4x- 1)) dx`
= `1/2 int(sqrt(4x - 1) - 13/sqrt(4x - 1)) dx`
= `1/2 int(4x- 1)^(1/2) "d"x - 13/2 int(4x - 1)^(1/2) dx`
= `1/2[((4x - 1)^(3/2))/(3/2) xx 1/4] - 13/2 [((4x - 1)^(1/2))/(1/2) xx 1/4]`
∴ `I = 1/12(4x - 1)^(3/2) - 13/4 sqrt(4x - 1) +c`
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