Integration

Let u = e^-st  then du/dt = -se^-st or e^-stdt = -(1/s)du so we can write the integral as:

$\int te^{-st}dt \rightarrow -\frac{1}{s}\int tdu \rightarrow -\frac{1}{s}\int td(e^{-st})$

Excellent, thank you for explaining.

Thanks Alan,

I would not have got that either without you showing me.

I just want to finish it.

Let

$u = e^{-st} \;\; then\;\; \\\frac{du}{dt} = -se^{-st} \\e^{-st}dt = \frac{-1}{s}du$

so we can write the integral as:

$\int te^{-st}dt \rightarrow -\frac{1}{s}\int tdu \rightarrow -\frac{1}{s}\int td(e^{-st})$

$u=e^{-st}\\ ln(u)=-st\\ t=\frac{-ln(u)}{s}$

So

$\frac{-1}{s}\int t\;du\\ =\frac{-1}{s}\int \frac{-ln(u)}{s}\;du\\ =\frac{1}{s^2}\int 1*ln(u)\;du\\ \qquad \text{integrate by parts}\\ \qquad \frac{da}{du}=1\quad b=ln(u)\\ \qquad a=u\quad \frac{db}{du}=\frac{1}{u}\\ =\frac{1}{s^2}\left( [ab]-\int a*\frac{db}{du}\:du\right)\\ =\frac{1}{s^2}\left( [uln(u)]-\int u*\frac{1}{u}\:du\right)\\ =\frac{1}{s^2}\left( [uln(u)]-\int 1\:du\right)\\ =\frac{1}{s^2}\left( [uln(u)]-u\right)+c\\ =\frac{u}{s^2}\left( ln(u)-1+k \right)$

Why would you ever want to present it with the extra step in the question? 

But u = e^(-st) so:

$\dfrac{u}{s^2}(\ln u-1)+k\\ =\dfrac{e^{-st}}{s^2}(-st-1)+k$