I've worked through a long substitution question and have come to this point.
t^2(du/dt) = 2te^(-u)
I know I have to integrate both sides but am struggling with how to do it?!
$\displaystyle \displaystyle t^2\frac{du}{dt} = 2te^{-u}$
Let's start by cancelling t from both sides,
$\displaystyle \displaystyle t\frac{du}{dt} = 2e^{-u}$
Now we'll isolate the derivative as this may be separable.
$\displaystyle \displaystyle \frac{du}{dt} = \frac{2e^{-u}}{t}$
Seems we can now separate so,
$\displaystyle \displaystyle \frac{du}{2e^{-u}} = \frac{dt}{t}$
Neatening up a little
$\displaystyle \displaystyle \frac{1}{2}e^{u}~du = \frac{dt}{t}$
And integrating
$\displaystyle \displaystyle \int \frac{1}{2}e^{u}~du = \int \frac{dt}{t}$
$\displaystyle \displaystyle \frac{1}{2}e^{u} = \ln (t)+C$
Finishing,
$\displaystyle \displaystyle e^{u} = 2\ln (t)+C$
$\displaystyle \displaystyle u = \ln{( 2\ln (t)+C)}$