Question

How do you find `intsqrtx lnxdx` using integration by parts?

`intsqrtx lnxdx`

The answer is supposedly `2/3 xsqrtx lnx - 4/9 xsqrtx + C`, but I don't understand how it came to be.

Please explain. Thanks in advance!

Answer 1

`intsqrtxlnx=2/3xsqrtxlnx-4/9xsqrtx+C`

Explanation:

Make the following selections:

`u=lnx`

Calculate its differential:

`(du)/dx=1/x -> du=dx/x`

`dv=sqrtxdx`

Integrate to find `v:`

`v=intsqrtxdx=intx^(1/2)dx=2/3x^(3/2)` (no constant of integration needed here, we put it in at the end)

Note that we selected the specific choices `u=lnx, v=sqrtxdx` as if we chose things the other way around, we'd need to calculate `v=intlnxdx` which requires using integration by parts -- again.

Apply the integration by parts formula:

`uv-intvdu=2/3x^(3/2)lnx-2/3intx^(3/2)/xdx`

`x^(3/2)/x=x^(3/2-1)=x^(1/2)`

Thus, we have

`2/3x^(3/2)lnx-2/3intx^(1/2)dx=2/3x^(3/2)lnx-(2/3)(2/3)x^(3/2)+C`

`intsqrtxlnx=2/3x^(3/2)lnx-4/9x^(3/2)+C`

Now, `x^(3/2)=x^(1/2)x=xsqrtx`:

`intsqrtxlnx=2/3xsqrtxlnx-4/9xsqrtx+C`