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Determine the integral of y=(x+3)/(x^2-9)? 

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minlux | Honors

Posted September 30, 2013 at 3:42 PM via web

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Determine the integral of y=(x+3)/(x^2-9)? 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 30, 2013 at 4:34 PM (Answer #1)

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You need to evaluate the given indefinite integral, hence, you should first reduce the integrand to the most simplified form, such that:

`(x + 3)/(x^2 - 9) = (x + 3)/((x - 3)(x + 3))`

Reducing the duplicate factors, yields:

`(x + 3)/(x^2 - 9) = 1/(x - 3)`

Hence, evaluating the indefinite integral, yields:

`int (x + 3)/(x^2 - 9) dx = int 1/(x - 3) dx`

You should come up with the following substitution, such that:

`x - 3 = t => dx = dt`

Replacing the variable, yields:

`int 1/(x - 3) dx = int 1/t dt`

`int 1/t dt = ln|t| + c`

Replacing back `x - 3` for t yields:

`int 1/(x - 3) dx = ln|x - 3| + c`

Hence, evaluating the given indefinite integral, yields
`int (x + 3)/(x^2 - 9) dx = ln|x - 3| + c` .

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