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

Answer 1

Elijah Abram

#2xlnx-2x+1/2x^2+C#

I'm assuming that you have: #lnx^2+x#.

Next up is:

#intlnx^2+x \ dx#

#=intlnx^2 \ dx+intx \ dx#

Using the fact that #log(a^b)=bloga# and evaluating the second integral, we get:

#=int2lnx \ dx+1/2x^2+c_1#

#=2intlnx \ dx+1/2x^2+c_1#

Let's evaluate #intlnx \ dx#. Use integration by parts:

#intu \ dv=uv-intv \ du#

Let #u=lnx,dv=1#.

#:.v=x,du=1/x#

When combined, we obtain:

#intlnx \ dx=xlnx-intx*1/x \ dx#

#=xlnx-int1 \ dx#

#=xlnx-x#

#=xlnx-x+c_2#

Thus, we obtain:

#=2(xlnx-x)+c_2+1/2x^2+c_1# (notice how I didn't put #2c_2# because constant ALWAYS goes after full integration)

#=2xlnx-2x+c_2+1/2x^2+c_1#

#=2xlnx-2x+1/2x^2+C#

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Answer 2

Christopher Agresta

The antiderivative of ln(x)^2 + x dx is (x^2 * ln(x))/2 - x^2/4 + 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.