How do you find intsqrtx lnxdx using integration by parts?

$\int \sqrt{x} \ln x \mathrm{dx}$ The answer is supposedly $\frac{2}{3} x \sqrt{x} \ln x - \frac{4}{9} x \sqrt{x} + C$, but I don't understand how it came to be. Please explain. Thanks in advance!

Apr 21, 2018

$\int \sqrt{x} \ln x = \frac{2}{3} x \sqrt{x} \ln x - \frac{4}{9} x \sqrt{x} + C$

Explanation:

Make the following selections:

$u = \ln x$

Calculate its differential:

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \to \mathrm{du} = \frac{\mathrm{dx}}{x}$

$\mathrm{dv} = \sqrt{x} \mathrm{dx}$

Integrate to find $v :$

$v = \int \sqrt{x} \mathrm{dx} = \int {x}^{\frac{1}{2}} \mathrm{dx} = \frac{2}{3} {x}^{\frac{3}{2}}$ (no constant of integration needed here, we put it in at the end)

Note that we selected the specific choices $u = \ln x , v = \sqrt{x} \mathrm{dx}$ as if we chose things the other way around, we'd need to calculate $v = \int \ln x \mathrm{dx}$ which requires using integration by parts -- again.

Apply the integration by parts formula:

$u v - \int v \mathrm{du} = \frac{2}{3} {x}^{\frac{3}{2}} \ln x - \frac{2}{3} \int {x}^{\frac{3}{2}} / x \mathrm{dx}$

${x}^{\frac{3}{2}} / x = {x}^{\frac{3}{2} - 1} = {x}^{\frac{1}{2}}$

Thus, we have

$\frac{2}{3} {x}^{\frac{3}{2}} \ln x - \frac{2}{3} \int {x}^{\frac{1}{2}} \mathrm{dx} = \frac{2}{3} {x}^{\frac{3}{2}} \ln x - \left(\frac{2}{3}\right) \left(\frac{2}{3}\right) {x}^{\frac{3}{2}} + C$

$\int \sqrt{x} \ln x = \frac{2}{3} {x}^{\frac{3}{2}} \ln x - \frac{4}{9} {x}^{\frac{3}{2}} + C$

Now, ${x}^{\frac{3}{2}} = {x}^{\frac{1}{2}} x = x \sqrt{x}$:

$\int \sqrt{x} \ln x = \frac{2}{3} x \sqrt{x} \ln x - \frac{4}{9} x \sqrt{x} + C$