# How do you integrate #int x/(sqrt(2x-1))dx# from [1,5]?

The answer is

Let's rewrite the function

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To integrate the function ( \frac{x}{\sqrt{2x-1}} ) from 1 to 5, you can use the substitution method. Let ( u = \sqrt{2x-1} ), then ( x = \frac{u^2 + 1}{2} ) and ( dx = \frac{u}{\sqrt{2x-1}} , du ).

Now, substitute these expressions into the integral:

[ \int_{1}^{5} \frac{x}{\sqrt{2x-1}} , dx = \int_{\sqrt{2(1)-1}}^{\sqrt{2(5)-1}} \frac{\frac{u^2+1}{2}}{u} \cdot \frac{u}{\sqrt{2x-1}} , du ]

[ = \int_{\sqrt{1}}^{\sqrt{9}} \frac{u^2+1}{2\sqrt{2u^2-1}} , du ]

[ = \int_{1}^{3} \frac{u^2+1}{2\sqrt{2u^2-1}} , du ]

This integral can be further simplified using algebraic manipulations and trigonometric substitution. Once you find the antiderivative, evaluate it at the upper and lower limits of integration (3 and 1, respectively) and subtract the lower limit result from the upper limit result 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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