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

Answer 1

I would set : #6-x=t^2# so: #x=6-t^2# and #dx=-2tdt# Substituting: #int(6-t^2)/t(-2t)dt=int(-12+2t^2)dt=-12t+2t^3/3+c# Going back to #x#: #=-12sqrt(6-x)+2/3(6-x)sqrt(6-x)+c#

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Answer 2

Charles Arocha

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.