What is integral (down 4-up 9) 1/(xsquare root x - 2x + square root x) ?

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You need to evaluate the following definite integral, such that:

`int_4^9 1/(xsqrt x - 2x + sqrt x)dx`

You should factor out `sqrt x` to denominator, such that:

`int_4^9 1/(sqrtx(x - 2sqrt x + 1))dx`

You should notice that `x - 2sqrt x + 1` represents the expansion of binomial `(sqrt x - 1)^2` , hence, replacing `(sqrt x - 1)^2` for `x - 2sqrt x + 1` , yields:

`int_4^9 1/(sqrtx*(sqrt x - 1)^2)dx`

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

`sqrt x = t => sqrt x - 1 = t - 1`

`1/(2sqrt x)dx = dt => dx = 2tdt`

You need to change the limits of integration, such that:

`sqrt 4 = 2 = t `

`sqrt 9 = 3 = t`

Changing the variable, yields:

`int_2^3 (2tdt)/(t*(t - 1)^2)`

Reducing the duplicate factors yields:

`int_2^3 (2dt)/((t - 1)^2) = 2 int_2^3 (dt)/(t - 1)^2`

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

`t - 1 = u => dt = du`

`2 - 1 = 1 = u`

`3 - 1 = 2 = u`

Changing the variable, yields:

`2 int_1^2 (du)/u^2 = -2/u|_1^2`

Using the fundamental formula of calculus, yields:

`2 int_1^2 (du)/u^2 = -2(1/2 - 1/1) = -2*(-1/2) = 1`

**Hence, evaluating the definite integral, using the indicated double substitutions, yields**` int_4^9 1/(xsqrt x - 2x + sqrt x)dx = 1.`

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