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

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Answer 1 · 2018-06-29T13:58:18

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

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

Then we have:

$intlnx^2+x \ dx$

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

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 $u=lnx,dv=1$ .

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

Putting together, we get:

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

$=xlnx-int1 \ dx$

$=xlnx-x$

$=xlnx-x+c_2$

So, we get:

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

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

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

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