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How do you integrate #((x)sqrt(x-1))dx#?

Answer 1

Ezra Adams

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Answer 2

Aurora Alvarado

To integrate (\int x \sqrt{x - 1} , dx), you can use the method of integration by parts. Let (u = x) and (dv = \sqrt{x - 1} , dx). Then, (du = dx) and (v = \frac{2}{3}(x - 1)^\frac{3}{2}). Applying the integration by parts formula, we have:

[ \int x \sqrt{x - 1} , dx = uv - \int v , du ]

Substituting the values of (u), (v), (du), and (dv), we get:

[ \int x \sqrt{x - 1} , dx = x \cdot \frac{2}{3}(x - 1)^\frac{3}{2} - \int \frac{2}{3}(x - 1)^\frac{3}{2} , dx ]

Now, integrate the second term:

[ \int \frac{2}{3}(x - 1)^\frac{3}{2} , dx = \frac{2}{3} \cdot \frac{2}{5}(x - 1)^\frac{5}{2} + C ]

Therefore, the integral becomes:

[ \int x \sqrt{x - 1} , dx = x \cdot \frac{2}{3}(x - 1)^\frac{3}{2} - \frac{4}{15}(x - 1)^\frac{5}{2} + C ]

Where (C) is the constant of integration.

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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.