# Math Help - Integration problem

1. ## Integration problem

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?!

2. $\displaystyle t^2\frac{du}{dt} = 2te^{-u}$

Let's start by cancelling t from both sides,

$\displaystyle t\frac{du}{dt} = 2e^{-u}$

Now we'll isolate the derivative as this may be separable.

$\displaystyle \frac{du}{dt} = \frac{2e^{-u}}{t}$

Seems we can now separate so,

$\displaystyle \frac{du}{2e^{-u}} = \frac{dt}{t}$

Neatening up a little

$\displaystyle \frac{1}{2}e^{u}~du = \frac{dt}{t}$

And integrating

$\displaystyle \int \frac{1}{2}e^{u}~du = \int \frac{dt}{t}$

$\displaystyle \frac{1}{2}e^{u} = \ln (t)+C$

Finishing,

$\displaystyle e^{u} = 2\ln (t)+C$

$\displaystyle u = \ln{( 2\ln (t)+C)}$