Question

How do you integrate `int (lnx)^2` by integration by parts method?

Answer 1

`x(lnx)^2-2xlnx+2x+C`

Explanation:

`I=int(lnx)^2dx`

Integration by parts takes the form: `intudv=uv-intvdu`

So, for the given integral, let:

`{(u=(lnx)^2" "=>" "du=(2lnx)/xdx),(dv=dx" "=>" "v=x):}`

Plugging these into the integration by parts formula, this becomes:

`I=x(lnx)^2-intx((2lnx)/x)dx`

`I=x(lnx)^2-2intlnxdx`

To solve this integral, we will reapply integration by parts:

`{(u=lnx" "=>" "du=1/xdx),(dv=dx" "=>" "v=x):}`

Thus, this becomes:

`I=x(lnx)^2-2[xlnx-intx(1/x)dx]`

`I=x(lnx)^2-2xlnx+2intdx`

`I=x(lnx)^2-2xlnx+2x+C`

Answer 2

`:. int (lnx)^2 dx = x(lnx)^2-2xlnx +2x +c`

Explanation:

Remember the formula for IBP: `int u(dv)/dxdx=uv-intv(du)/dxdx`

Let `u=lnx=>(du)/dx=1/x`
Let `(dv)/dx=lnx=>v=intlndx=xlnx - x` (see additional notes)

Substitute into the IBP equation:
`int (lnx)(lnx) dx = (lnx)(xlnx-x) - int (xlnx-x)1/x dx`
`:. int (lnx)^2 dx = x(lnx)^2-xlnx - int (lnx-1 )dx`
`:. int (lnx)^2 dx = x(lnx)^2-xlnx - (xlnx-x -x) +c`
`:. int (lnx)^2 dx = x(lnx)^2-xlnx - xlnx +2x +c`
`:. int (lnx)^2 dx = x(lnx)^2-2xlnx +2x +c`

Additional Notes:
How do you find `intlndx=xlnx - x`. Firstly its is extremely helpful to learn this result, but if you can't or you need to prove it then you need to use IBP again:

Let `u=lnx=>(du)/dx=1/x`
Let `(dv)/dx=1=>v=x`
Substitute into the IBP equation:
`int (lnx)(1)dx=(lnx)(x)-int(x*1/x)dx`
`:. int lnxdx=xlnx-int(1)dx`
`:. int lnxdx=xlnx-x` `(+c)`