Question

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

Answer 1

`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`
...........................................................................................................
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`

Answer 2

`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`