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प्रश्न
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
उत्तर
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
` = int "dx"/sqrt(2[5/2-2"x"-"x"^2]`
` =1/sqrt2int"dx"/sqrt(5/2 - 2"x" - "x"^2)`
` = 1/sqrt2 int"dx"/sqrt(5/2-("x"^2+2"x"))`
` = 1/sqrt2 int"dx"/sqrt(5/2 -("x"^2+2"x"+1-1))`
` = 1/sqrt2 int"dx"/sqrt(5/2 -("x"+1)^2+1`
` = 1/sqrt2 int"dx"/(7/2-("x"+1)^2)`
` = 1/sqrt2 int "dx"/sqrt((sqrt7/sqrt2)^2 - ("x"+1)^2)`
` = 1/sqrt2sin^-1((("x"+1)sqrt2)/sqrt7) + "C"`
` = 1/sqrt2sin^-1 (sqrt(2/7) ("x"+1)) + "C"`
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