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

2 Answers
Jun 14, 2018

Answer:

#I=2x[lnx-1]+c#

Explanation:

Here,

#I=intln(x)^2dx#

Using Power-coefficient Rule :

#color(red)(lnx^n=n*lnx# , we get

#I=intlnx^2dx=int2*lnx...to(A)#

Using , Integration by parts:

#color(blue)(int(u*v)dx=uintvdx-int(u'intvdx)dx#

#u=lnx and v=2=>u'=1/x and intvdx=2x#

#I=lnx*2x-int1/x*2xdx+c#

#=>I=2x*lnx-int2dx+c#

#=>I=2xlnx-2x+c#

#=>I=2x[lnx-1]+c#
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Note:

If #I=(lnx)^2# ,then

#I=(lnx)^2*1dx#

Using , Integration by parts:

#u=(lnx)^2 and v=1=>u'=(2lnx)/x and intvdx=x#

#I=(lnx)^2int1dx-int(2lnx)/x*xdx#

#=(lnx)^2*x-int2*lnx#

Using above result for #(A)#

#I=x*(lnx)^2-{2x[lnx-1]}+c#

Jun 14, 2018

Answer:

#2xlnx-x+C#

Explanation:

The antiderivative of a function is basically the function's integral. So we get:

#intln(x)^2 \ dx#

I'm assuming that we have #intlnx^2 \ dx#.

Using logarithm rules, we get:

#=int2lnx \ dx#

#=2intlnx \ dx#

This is a common integral, where #intlnx \ dx=xlnx-x+C#.

#:.=2xlnx-2x+C#