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Compute the indefinite integral sign x+1/square root(x^2+2x+7)dx?

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mozostax | Student, Undergraduate | eNoter

Posted September 2, 2012 at 6:27 AM via web

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Compute the indefinite integral sign x+1/square root(x^2+2x+7)dx?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 2, 2012 at 6:43 AM (Answer #1)

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You need to use substitution method, hence, you should substitute y for `x^2+2x+7`  such that:

`x^2+2x+7 = y => (2x + 2)dx = dy => 2(x+1)dx = dy => (x+1)dx = (dy)/2`

Changing the variable yields:

`int (x+1)/sqrt(x^2+2x+7) dx = int (dy)/(2sqrty)`

Notice that `(sqrt y)' = 1/(2sqrty)` , hence `int (dy)/(2sqrty) = int (sqrt y)' dy`  such that:

`int (sqrt y)' dy = sqrt y + c`

Substituting `x^2+2x+7`  for y yields:

`int (x+1)/sqrt(x^2+2x+7) dx = sqrt(x^2+2x+7) + c`

Hence, evaluating the indefinite integral using substitution method yields `int (x+1)/sqrt(x^2+2x+7) dx = sqrt (x^2+2x+7) + c.`

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