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How do you integrate #int ln 4x dx # using integration by parts?

Answer 1

Elijah Abram

The answer is #=x(ln(|4x|)-1)+C#

Perform the integration by parts

#intuv'=uv-intu'v#

Here,

#u=ln4x#, #=>#, #u'=1/(4x)*4=1/x#

#v'=1#, #=>#, #v=x#

Therefore,

#intln4x=xln4x-int1/x*xdx#

#=xln4x-x+C#

#=x(ln(|4x|)-1)+C#

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Answer 2

Paisley Johnson

#=xln4x-x+C#

#intln4xdx#

You can't integrate #ln4x# so you pick

#u=ln4x# #rarr# #du=1/xdx# #dv=dx##rarr##v=x#

And by using the formula: #intudv=uv-intvdu#

#intln4xdx=xln4x-intx/xdx#

#=xln4x-int1dx#

#=xln4x-x+C#

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Answer 3

Samantha Adkison

To integrate ( \int \ln(4x) , dx ) using integration by parts, you would choose ( u = \ln(4x) ) and ( dv = dx ), then apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

First, differentiate ( u ) to find ( du ), and integrate ( dv ) to find ( v ). Then substitute these into the integration by parts formula and solve.

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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.