# How do I evaluate the integral #intx/(sqrt(6-x)) dx#?

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To evaluate the integral ∫x/√(6-x) dx:

- Perform a substitution: Let u = 6 - x. Then, du = -dx.
- Rewrite the integral in terms of u: ∫-x/√u du.
- Factor out the negative sign and rewrite x in terms of u: ∫x/√u du.
- Replace x with 6 - u.
- Integrate with respect to u.
- Substitute back u with 6 - x in the final result.

The integral should then be evaluated as follows:

= -∫(6 - u)/√u du = -∫(6/u - 1) du = -6∫(1/u) du + ∫1 du = -6ln|u| + u + C = -6ln|6 - x| + (6 - x) + C

So, the integral of x/√(6-x) dx is -6ln|6 - x| + (6 - x) + C.

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