We'll write the indefinite
integral:
Int 3dx/sqrt x = 3Int dx/sqrt
x
We'll multiply both, denominator and numerator, by
2:
2/2sqrtx = 2*(1/2sqrt x) =
2*(sqrtx)'
3Int dx/sqrt x = 3Int
2*(sqrtx)'dx
3Int 2*(sqrtx)'dx = 3*2Int
(sqrtx)'dx
3*2Int (sqrtx)'dx = 6 sqrt x +
C
Int 3dx/sqrt x = 6 sqrt x +
C
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