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How do you integrate # ln(sqrt(x)) #?

Answer 1

Charles Arocha

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

#int \ ln(sqrt(x)) \ dx#

using IBP: #int uv' = uv - int u'v#

#= int \ ln(sqrt(x)) * (x)' \ dx#

#= ln(sqrt(x)) * x - int (ln(sqrt(x)))' * x\ dx#

#= 1/2 x ln x - int \ 1/(sqrt(x)) (1/2 * 1/sqrtx) * x\ dx#

#= 1/2 x ln x - 1/2 int \ dx #

#= 1/2 x ln x - 1/2 x + C#

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

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Answer 2

Emilia Aguilar

To integrate ( \ln(\sqrt{x}) ), you can use substitution. Let ( u = \sqrt{x} ), then ( x = u^2 ) and ( dx = 2u du ). The integral becomes ( \int \ln(u) \cdot 2u , du ). Using integration by parts, with ( dv = 2u , du ) and ( u = \ln(u) ), you get ( v = u^2 ) and ( du = 1/u , du ).

The integral becomes ( uv - \int v , du ), which equals ( u^2 \ln(u) - \int u^2 \cdot \frac{1}{u} , du ). Simplifying, you get ( u^2 \ln(u) - \int u , du ), which integrates to ( u^2 \ln(u) - \frac{1}{2}u^2 + C ). Substituting back ( u = \sqrt{x} ), the result is ( x\ln(\sqrt{x}) - \frac{1}{2}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.