# How do you integrate #int 2/sqrt(3x-7)# using substitution?

Your equation becomes:

You can then substitute u into the square root and du for 3dx using the three on the top. You should then get

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To integrate ( \int \frac{2}{\sqrt{3x - 7}} ) using substitution, we can let ( u = 3x - 7 ). Then, find ( du ) by differentiating both sides with respect to ( x ), which gives ( du = 3dx ).

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

[ \int \frac{2}{\sqrt{3x - 7}} , dx ]

Let ( u = 3x - 7 ) and ( du = 3dx ), so ( \frac{du}{3} = dx ).

Substitute ( u ) and ( \frac{du}{3} ) into the integral:

[ \int \frac{2}{\sqrt{u}} \cdot \frac{du}{3} ]

This simplifies to:

[ \frac{2}{3} \int \frac{1}{\sqrt{u}} , du ]

Now integrate ( \frac{1}{\sqrt{u}} ) with respect to ( u ):

[ \frac{2}{3} \int u^{-1/2} , du ]

Using the power rule for integration, we get:

[ \frac{2}{3} \cdot \frac{u^{1/2}}{1/2} + C ]

Simplifying further:

[ \frac{4}{3} \sqrt{u} + C ]

Finally, substitute back ( u = 3x - 7 ) to get the result in terms of ( x ):

[ \frac{4}{3} \sqrt{3x - 7} + C ]

So, the integral ( \int \frac{2}{\sqrt{3x - 7}} , dx ) evaluates to ( \frac{4}{3} \sqrt{3x - 7} + 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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