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!

1 Answer
Apr 21, 2018

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