# How do you integrate #int (sqrtx-1)^2/sqrtx# using substitution?

The answer is

Use the substitution

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To integrate (\int \frac{(\sqrt{x} - 1)^2}{\sqrt{x}} dx), use substitution.

Let (u = \sqrt{x}), which means (x = u^2). Consequently, (dx = 2u du).

Now, rewrite the integral in terms of (u):

[ \int \frac{(u - 1)^2}{u} \cdot 2u du ]

This simplifies to:

[ 2\int (u - 1)^2 du ]

Expanding ((u - 1)^2) gives:

[ 2\int (u^2 - 2u + 1) du ]

Integrating term by term:

[ 2\left(\frac{u^3}{3} - u^2 + u\right) + C ]

Since (u = \sqrt{x}), substitute back to (x):

[ \frac{2}{3}x^{3/2} - 2x + 2\sqrt{x} + C ]

So, the integral (\int \frac{(\sqrt{x} - 1)^2}{\sqrt{x}} dx) is:

[ \frac{2}{3}x^{3/2} - 2x + 2\sqrt{x} + C ]

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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.

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.

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.

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