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What is the antiderivative of #ln(x)^2#?

Answer 1

# int (lnx)^2dx = x(lnx)^2 +2x +C#

You should learn the IBP formula: # int u(dv)/dxdx=uv - int v (du)/dxdx #

So essentially we are looking for one function that simplifies when it is differentiated, and one that simplifies when integrated (or at least is integrable).

We also need to know that #int lnxdx=xlnx-x# (either learn or use IBP):

Let # {(u=lnx, => ,(du)/dx=1/x),((dv)/dx=lnx,=>,v =xlnx-x ):}#

Then IBP gives; # int lnx lnx dx=lnx(xlnx-x) - int (xlnx-x)1/xdx # # :. int (lnx)^2 dx = x(lnx)^2 - xlnx - int (lnx-1)dx # # :. int (lnx)^2 dx = x(lnx)^2 - xlnx - (xlnx-x-x) # # :. int (lnx)^2 dx = x(lnx)^2 - xlnx + xlnx +2x #

And so we have: # :. int (lnx)^2dx = x(lnx)^2 +2x +C#

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Answer 2

Scarlett Allison

The antiderivative of ln(x)^2 is ∫ln(x)^2 dx = x*ln(x)^2 - 2∫ln(x) dx.

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Answer 3

Charles Anctil

The antiderivative of ( \ln(x)^2 ) is ( x \ln(x)^2 - 2\int \ln(x) , dx ).

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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.