MHB What is the Integration by Parts method used for?

[karush]

$\tiny\text{Whitman 8.7.26 Integration by Parts} $ nmh{818}
$\displaystyle
I=\int t
\left(\ln\left({t}\right) \right)^2
\ d{t}
=\frac{{t}^{2}\left(\ln\left({t}\right)\right)^2 }{2}
-\frac{{t}^{2}\left(\ln\left({t}\right)\right) }{2}
+\frac{{t}^{2}}{4}
+ C$
$\displaystyle uv-\int v \ d{u} $

$\begin{align}\displaystyle
u& = t &
dv&=\left(\ln\left({t}\right) \right)^2 \ d{t} \\
du&= dt&
v& =t\left(\ln\left({t}\right)^2 - 2\ln\left({t}\right)+2\right)
\end{align}$
$\text{$v$ reintroduced $t$
so not sure of $u$ $dv$ substitutions } $

 

[MarkFL]

According to LIATE, you should first try:

$$u=\ln^2(t)\,\therefore\, du=2\ln(t)\frac{1}{t}\,dt$$

$$dv=t\,dt\,\therefore\,v=\frac{t^2}{2}$$

 

[karush]

$\tiny\text{Whitman 8.7.26 Integration by Parts} $
$\displaystyle
I=\int t
\left(\ln\left({t}\right) \right)^2
\ d{t}
=\frac{{t}^{2}\left(\ln\left({t}\right)\right)^2 }{2}
-\frac{{t}^{2}\left(\ln\left({t}\right)\right) }{2}
+\frac{{t}^{2}}{4}
+ C$
$\displaystyle uv-\int v \ d{u} $

$\begin{align}\displaystyle
u& = \ln^2 \left({t}\right) &
dv&= t \ dt \\
du&=\frac{2\ln\left({t}\right)}{t} \ dt&
v& =\frac{t^2}{2}
\end{align}$
So
$\displaystyle
\frac{{t}^{2}\ln^2 \left({t}\right)}{2}
-\int\frac{t^2}{2} \frac{2\ln\left({t}\right)}{t} \ dt
\implies\frac{{t}^{2}\ln^2 \left({t}\right)}{2}
-\int t \ln\left({t}\right) \ dt $
For
$\displaystyle
\int t \ln\left({t}\right) \ dt $

$\begin{align}\displaystyle
u& = \ln\left({t}\right) &
dv&= t \ dt \\
du&=\frac{1}{t} \ dt&
v& =\frac{t^2}{2}
\end{align}$

$\displaystyle\frac{{t}^{2}\ln\left({t}\right)}{2}
-\int\frac{t}{2} \ dt =\frac{{t}^{2}\ln\left({t}\right)}{2}- \frac{{t}^{2}}{4}$
and finally...
$\displaystyle
\frac{{t}^{2}\left(\ln\left({t}\right)\right)^2 }{2}
-\frac{{t}^{2}\left(\ln\left({t}\right)\right) }{2}
+\frac{{t}^{2}}{4}
+ C$