Integral of t*e^(-t)*dt dx

1. Let $u = - t$.

   Then let $du = - dt$ and substitute $du$:

   $$\int u e^{u}\, du$$

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

      Then $\operatorname{du}{\left(u \right)} = 1$.

      To find $v{\left(u \right)}$:

      1. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      Now evaluate the sub-integral.

   2. The integral of the exponential function is itself.

      $$\int e^{u}\, du = e^{u}$$

   Now substitute $u$ back in:

   $$- t e^{- t} - e^{- t}$$

2. Now simplify:

   $$- \left(t + 1\right) e^{- t}$$

3. Add the constant of integration:

   $$- \left(t + 1\right) e^{- t}+ \mathrm{constant}$$

The answer is:

$$- \left(t + 1\right) e^{- t}+ \mathrm{constant}$$