# How do you find the integral of sqrt (t)*ln (t) dt?

Sep 4, 2015

use integration by parts to find $\int \sqrt{t} \ln \left(t\right) \mathrm{dt} = \frac{2}{3} {t}^{\frac{3}{2}} \left(\ln \left(t\right) - \frac{2}{3}\right) + C$

#### Explanation:

We use integration by parts by noting that $\sqrt{t} = \frac{d}{\mathrm{dt}} \left(\frac{2}{3} {t}^{\frac{3}{2}}\right)$.
Therefore
$\int \sqrt{t} \ln \left(t\right) \mathrm{dt} = \frac{2}{3} \int \ln \left(t\right) d \left({t}^{\frac{3}{2}}\right) = \frac{2}{3} \left(\ln \left(t\right) {t}^{\frac{3}{2}} - \int {t}^{\frac{3}{2}} \mathrm{dl} n \left(t\right)\right)$.
Since $\frac{d}{\mathrm{dt}} \left(\ln \left(t\right)\right) = \frac{1}{t}$ this gives
$\int \sqrt{t} \ln \left(t\right) \mathrm{dt} = \frac{2}{3} \left(\ln \left(t\right) {t}^{\frac{3}{2}} - \int {t}^{\frac{3}{2}} / t \mathrm{dt}\right) = \frac{2}{3} \left(\ln \left(t\right) {t}^{\frac{3}{2}} - \int {t}^{\frac{1}{2}} \mathrm{dt}\right) = \frac{2}{3} \left(\ln \left(t\right) {t}^{\frac{3}{2}} - \frac{2}{3} {t}^{\frac{3}{2}}\right) + C = \frac{2}{3} {t}^{\frac{3}{2}} \left(\ln \left(t\right) - \frac{2}{3}\right) + C$
With C the integration constant.