# How do you integrate #int sqrt(1+2x)# using substitution?

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How do you integrate #int sqrt(1+2x)dx# using substitution?

How do you integrate

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To integrate (\int \sqrt{1 + 2x} , dx) using substitution:

- Let (u = 1 + 2x), then (du = 2 , dx).
- Solve for (dx) to substitute back into the integral: (dx = \frac{du}{2}).
- Substitute (u) and (dx) into the integral: [\int \sqrt{u} \cdot \frac{du}{2}]
- Simplify and pull out constants: [\frac{1}{2} \int \sqrt{u} , du]
- Integrate (\sqrt{u}): [\frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} + C]
- Substitute back (u = 1 + 2x): [\frac{1}{3} (1 + 2x)^{\frac{3}{2}} + C]

So, (\int \sqrt{1 + 2x} , dx = \frac{1}{3} (1 + 2x)^{\frac{3}{2}} + C), where (C) is the constant of integration.

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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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