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What is the indefinite integral of #ln(x^3)/x#?

Answer 1

#int(ln(x^3))/xdx=(3(lnx)^2)/2+C#

We'll start by making life a lot easier and using a property of the natural log to simplfy things: #ln(x^a)=alnx#

Utilizing this property, #int(ln(x^3))/xdx# becomes #int(3lnx)/xdx#, and furthermore, since 3 is constant, #3intlnx/xdx#.

Now, notice that we have #lnx# and its derivative, #1/x#. This makes the integral a textbook case of a #u#-substitution: #u=lnx->(du)/dx=1/x->du=1/xdx#

We can rewrite the integral a little to make it easier to follow along: #3int(lnx)(1/x)dx#

Because #color(red)u=lnx# and #color(blue)(du)=1/xdx#, #3intcolor(red)(lnx)color(blue)((1/x)dx)=3intcolor(red)ucolor(blue)(du)#

Our new integral is easily evaluated by the reverse power rukle: #3intudu=3(u^2/2+C)# #color(white)(XX)=(3u^2)/2+C#

Finally, because #u=lnx#, #(3u^2)/2+C=(3(lnx)^2)/2+C#

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Answer 2

Christopher Agresta

The indefinite integral of ( \frac{\ln(x^3)}{x} ) is ( \frac{1}{2} \ln(x^3)^2 + C ), where ( C ) is the constant of integration.

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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.