integration 3x+2/cubic root x^2+2x+5   (please solve it)solve integration

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should come up with the substitution `x^2+2x+5 = y` , hence, differentiating both sides yields:

`(2x + 2) dx = dy`

You need to write the function in terms of y such that:

`int ((2x + 2) dx)/(root(3)(x^2+2x+5)) = int (dy)/(root(3)(y))`

You should write the cube root such that `root(3)(y) = y^(1/3), ` hence `1/y^(1/3) = y^(-1/3)`

`int (dy)/(root(3)(y)) = int (y^(-1/3)dy)`

`int (y^(-1/3)dy) = y^(-1/3 + 1)/(-1/3 + 1) + c`

`int (y^(-1/3)dy) =y^(2/3)/(2/3) + c`

`int (y^(-1/3)dy) = 3y^(2/3)/2 + c`

You need to substitute `x^2+2x+5`  for y such that:

int ((2x + 2) dx)/(root(3)(x^2+2x+5)) = 3(root(3)(x^2+2x+5)^2)/2 + c

Hence, integrating `(2x + 2)/(root(3)(x^2+2x+5)) ` yields `int ((2x + 2) dx)/(root(3)(x^2+2x+5)) = 3(root(3)(x^2+2x+5)^2)/2 + c.`

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