# How do you find the definite integral of #x/sqrt(2x+1)# from #[0,4]#?

Use substitution to evaluate

Substitute:

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To find the definite integral of ( \frac{x}{\sqrt{2x+1}} ) from 0 to 4, you can use the substitution method. Let ( u = \sqrt{2x+1} ), then ( x = \frac{u^2 - 1}{2} ) and ( du = \frac{1}{\sqrt{2x+1}} dx ).

After substitution, the integral becomes ( \int_{1}^{3} \frac{u^2 - 1}{2u} du ). Simplify the integrand to ( \frac{1}{2} \int_{1}^{3} (u - \frac{1}{u}) du ).

Integrate term by term: ( \frac{1}{2} \left[ \frac{u^2}{2} - \ln|u| \right]_{1}^{3} ).

Substitute back for ( u ): ( \frac{1}{4} \left[ (3^2 - 1) - \ln(3) - (1^2 - 1) + \ln(1) \right] ).

Simplify: ( \frac{1}{4} \left[ 8 - \ln(3) \right] ).

So, the definite integral of ( \frac{x}{\sqrt{2x+1}} ) from 0 to 4 is ( \frac{1}{4} \left[ 8 - \ln(3) \right] ).

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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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