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प्रश्न
Integrate the following with respect to x.
`sqrt(3x + 5)`
योग
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उत्तर
`sqrt(3x + 5) = (3 + 5)^(1/2)`
This is of the form (ax + b)n where a = 3, b = 5, n = `1/2`
`int ("a"x + "b")^"n" "d"x = ("a"x + "b")^("n" + 1)/("a"("n" + 1)) + "c"`
So `int (3x + 5)^(1/2) "d"x = (3x + 5)^(1/2 + 1)/(3(1/2 + 1))`
= `(3x + 5)^(3/2)/(3(3/2)`
= `(3x + 5)^(3/2)/(9/2) + "c"`
= `2/9 (3x + 5)^(3/2) + "c"`
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