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

Answer:

#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#