# How do you integrate t^(1/2)*ln(t)?

##### 1 Answer
Mar 5, 2018

$I = \frac{2}{9} \cdot {t}^{\frac{3}{2}} \left[3 \cdot \ln t - 2\right] + C$

#### Explanation:

Integration by Parts:
$\textcolor{red}{\int u \cdot v \mathrm{dt} = u \int v \mathrm{dt} - \int \left\{\frac{\mathrm{du}}{\mathrm{dt}} \int v \mathrm{dt}\right\} \mathrm{dt}}$
Let,$u = \ln t , \mathmr{and} , v = {t}^{\frac{1}{2}} \Rightarrow \frac{\mathrm{du}}{\mathrm{dt}} = \frac{1}{t} , \int v \mathrm{dt} = {t}^{\left(\frac{1}{2}\right) + 1} / \left(\left(\frac{1}{2}\right) + 1\right) = \frac{t \left(\frac{3}{2}\right)}{\frac{3}{2}}$
$I = \int {t}^{\frac{1}{2}} \cdot \ln t \mathrm{dt} = \ln t \cdot \int {t}^{\frac{1}{2}} \mathrm{dt} - \int \frac{d}{\mathrm{dt}} \left(\ln t\right) \mathrm{dt} \cdot \int {t}^{\frac{1}{2}} \mathrm{dt} \Rightarrow I = \ln t \cdot {t}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - \int \frac{1}{t} \cdot {t}^{\frac{3}{2}} / \left(\frac{3}{2}\right) \cdot \mathrm{dt} = \frac{2}{3} \cdot {t}^{\frac{3}{2}} \ln t - \frac{2}{3} \int {t}^{\frac{3}{2} - 1} \mathrm{dt} = \frac{2}{3} {t}^{\frac{3}{2}} \cdot \ln t - \frac{2}{3} \int {t}^{\frac{1}{2}} \mathrm{dt} = \frac{2}{3} {t}^{\frac{3}{2}} \cdot \ln t - \frac{2}{3} \cdot {t}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C$
$I = \frac{2}{3} \cdot {t}^{\frac{3}{2}} \cdot \ln t - \frac{4}{9} \cdot {t}^{\frac{3}{2}} + C$
$I = \frac{2}{9} \cdot {t}^{\frac{3}{2}} \left[3 \cdot \ln t - 2\right] + C$