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How do you integrate #(5/sqrt x) dx#?

Answer 1

Samantha Adkison

#int 5/(sqrt(x))d x=10 sqrt x +C#

#int 5/(sqrt(x))d x=?#

#"we can rearrange the term as;" #

#int 5(x)^(-1/2)d x=5 int(x)^(-1/2) d x#

#int int 5/(sqrt(x))d x=5(1/(-1/2+1)) x^(-1/2+1)#

#int 5/(sqrt(x))d x=5*2*x^(1/2)#

#int 5/(sqrt(x))d x=10 sqrt x +C#

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Answer 2

Elijah Abram

To integrate ( \frac{5}{\sqrt{x}} , dx ), you can use the method of substitution. Let ( u = \sqrt{x} ). Then, ( du = \frac{1}{2\sqrt{x}} , dx ). Rearrange the equation to solve for ( dx ) to get ( dx = 2u , du ). Now substitute ( u ) and ( dx ) into the integral. The integral becomes ( \int \frac{5}{u} \times 2u , du ). Simplify to get ( \int 10 , du ). Finally, integrate ( 10 , du ) to get ( 10u + C ). Substitute back ( u = \sqrt{x} ) to get the final answer ( 10\sqrt{x} + C ).

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Answer from HIX Tutor

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.