# How do you integrate #int (2x^(3/2)+1)sqrtx# using substitution?

Calculate its differential:

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To integrate ( \int (2x^{3/2} + 1)\sqrt{x} ) using substitution, follow these steps:

- Let ( u = \sqrt{x} ).
- Find ( du ) by differentiating ( u ) with respect to ( x ).
- Express the integral in terms of ( u ) using the substitution ( u = \sqrt{x} ) and ( du ).
- Integrate with respect to ( u ).
- Finally, substitute back the expression involving ( x ).

So, let ( u = \sqrt{x} ), then ( du = \frac{1}{2\sqrt{x}} dx ). Rearranging this, we get ( dx = 2u , du ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int (2x^{3/2} + 1)\sqrt{x} , dx = \int (2u^3 + 1) \cdot u \cdot 2u , du = \int (4u^5 + 2u) , du ]

Integrate ( (4u^5 + 2u) ) with respect to ( u ) to get ( \frac{4}{6}u^6 + u^2 + C ), where ( C ) is the constant of integration.

Substitute back ( u = \sqrt{x} ) to get the final result:

[ \frac{4}{6}(\sqrt{x})^6 + (\sqrt{x})^2 + C = \frac{2}{3}x^{3/2} + x + C ]

Therefore, the integral of ( (2x^{3/2} + 1)\sqrt{x} ) with substitution is ( \frac{2}{3}x^{3/2} + 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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