# How do you use substitution to integrate #xsqrt(2x-1)#?

#int xsqrt(2x-1)dx=int (u+1)/2*sqrtu*(du)/2=1/4 int (usqrtu+sqrtu)du=
1/4*2/15*u^(3/2)*(3u+5)+c=1/15*(2x-1)^(3/2)*(3x+1)+c#

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To integrate ( x\sqrt{2x - 1} ) using substitution, follow these steps:

- Let ( u = 2x - 1 ).
- Find ( \frac{du}{dx} ).
- Rearrange the equation to solve for ( dx ).
- Substitute ( u ) and ( dx ) in terms of ( x ) into the integral.
- Integrate the resulting expression with respect to ( u ).
- Substitute back ( u ) in terms of ( x ) to find the final answer.

Let's go through the steps:

- Let ( u = 2x - 1 ).
- ( \frac{du}{dx} = 2 ).
- Rearranging, ( dx = \frac{du}{2} ).
- Substitute ( u ) and ( dx ) into the integral:

( \int x\sqrt{2x - 1} , dx = \int x\sqrt{u} \cdot \frac{du}{2} ) - Integrate the resulting expression with respect to ( u ):

( \frac{1}{2} \int x\sqrt{u} , du ) - Use integration by parts or another suitable method to integrate ( x\sqrt{u} ).
- Substitute back ( u ) in terms of ( x ) to find the final answer.

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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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