# How do you integrate int tln(t+1) by parts?

Nov 18, 2016

$I = \frac{1}{2} {t}^{2} \ln \left(t + 1\right) - \frac{1}{4} {t}^{2} - \frac{1}{2} t + \frac{1}{2} \ln | t + 1 | + C$

#### Explanation:

integration by parts formula

$\int u \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt} = u v - \int v \frac{\mathrm{du}}{\mathrm{dt}} \mathrm{dt}$

note function not defined for $t = - 1$

for

$\int \left(t \ln \left(t + 1\right)\right) \mathrm{dt}$

note function not defined for $t = - 1$

$u = \ln \left(t + 1\right) \implies \frac{\mathrm{du}}{\mathrm{dt}} = \frac{1}{t + 1}$

$\frac{\mathrm{dv}}{\mathrm{dt}} = t \implies v = \frac{1}{2} {t}^{2}$

$\therefore \int u \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt} = \frac{1}{2} {t}^{2} \ln \left(t + 1\right) - \frac{1}{2} \int {t}^{2} / \left(t + 1\right) \mathrm{dt}$

divide the improper fraction out

$\therefore \int u \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt} = \frac{1}{2} {t}^{2} \ln \left(t + 1\right) - \frac{1}{2} \int \left(\left(t - 1\right) + \frac{1}{t + 1}\right) \mathrm{dt}$

$\therefore \int u \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt} = \frac{1}{2} {t}^{2} \ln \left(t + 1\right) - \frac{1}{2} \left(\frac{1}{2} {t}^{2} - t + \ln | t + 1 |\right) + C$

$I = \frac{1}{2} {t}^{2} \ln \left(t + 1\right) - \frac{1}{4} {t}^{2} - \frac{1}{2} t + \frac{1}{2} \ln | t + 1 | + C$