# How do you integrate #(5x)/ (x-2)^(1/2) dx#?

Substitute back

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To integrate ( \frac{5x}{\sqrt{x - 2}} ), you can use the substitution method. Let ( u = x - 2 ). Then ( du = dx ). The integral becomes ( \int \frac{5(u + 2)}{\sqrt{u}} du ). This simplifies to ( \int 5u^{1/2} + 10u^{-1/2} du ). Now, integrate each term separately to get ( \frac{10}{3}u^{3/2} + 20u^{1/2} + C ). Finally, substitute back ( x - 2 ) for ( u ) to get ( \frac{10}{3}(x - 2)^{3/2} + 20\sqrt{x - 2} + C ), where ( C ) is the constant of integration.

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To integrate ( \frac{5x}{\sqrt{x - 2}} , dx ), you can use a substitution method. Let ( u = x - 2 ), then ( du = dx ).

[ \int \frac{5x}{\sqrt{x - 2}} , dx = \int \frac{5(u + 2)}{\sqrt{u}} , du ]

[ = \int 5(u^{\frac{1}{2}} + 2u^{-\frac{1}{2}}) , du ]

[ = 5 \left( \frac{2}{3} u^{\frac{3}{2}} + 4u^{\frac{1}{2}} \right) + C ]

[ = \frac{10}{3} (x - 2)^{\frac{3}{2}} + 20\sqrt{x - 2} + C ]

So, ( \int \frac{5x}{\sqrt{x - 2}} , dx = \frac{10}{3} (x - 2)^{\frac{3}{2}} + 20\sqrt{x - 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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