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How do you evaluate the integral #int sqrtx/(x-1)#?

Answer 1

Emma Aldridge

#= 2sqrtx + ln( (sqrtx-1) /(sqrtx +1) )+ C#

#int sqrtx/(x-1) dx#

it looks a little easier with a simplification: #y = sqrt x, dy = 1/(2 sqrt x) dx implies dx = 2 y dy#

giving us

#2 int ( y^2)/(y^2-1) dy#

#=2 int ( y^2 - 1 + 1)/(y^2-1) dy#

#=2 int 1 + ( 1)/(y^2-1) dy =2 int 1 + ( 1)/((y-1)(y +1)) dy#

#=2 int 1 + 1/2( 1/(y-1) - 1/(y +1)) dy#

#= 2y + ln( (y-1) /(y +1) )+ C#

#= 2sqrtx + ln( (sqrtx-1) /(sqrtx +1) )+ C#

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Answer 2

Ezekiel Alexander

To evaluate the integral ∫(√x)/(x - 1), you can use the method of substitution. Let u = √x, then du = (1/(2√x)) dx. Now substitute u and du into the integral:

∫(√x)/(x - 1) dx = ∫(u)/(u^2 - 1) * 2u du

This simplifies to:

∫(2u^2)/(u^2 - 1) du

Now you can use partial fraction decomposition to break down the integrand into simpler fractions. After that, integrate each term individually.

Once you find the antiderivative, don't forget to back-substitute u = √x to obtain the final result in terms of x.

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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.