Find the convolution of $x(t)*h(t)$ I am studying for an exam and have the following question:
$$x(t) = u(t)$$
$$h(t) = [e^{-t}-e^{-2t}]u(t)$$
where u(t) is a unit-step function. I need to find the convolution x(t)*h(t).
So:
$$
x(t)*h(t) = \int_{-\infty}^\infty u(t)[e^{-t}-e^{-2t}]u(t-\tau)d\tau\
$$
$$
x(t)*h(t) = u(t)[e^{-t}-e^{-2t}]\int_{-\infty}^\infty u(t-\tau)d\tau\
$$
The question I have is how do I take the integral of 
$$
\int_{-\infty}^\infty u(t-\tau)d\tau\
$$
And how would I graph this convolution? TIA!

using your notation, 
\begin{align*}
(x\ast h)(t) &= \int_\mathbb{R} u(\tau)h(t-\tau)d\tau  \overset{\textrm{u is step function}}{=} \int_0^\infty h(t-\tau)d\tau \\ &= \int_0^\infty (e^{-(t-\tau)}-e^{-2(t-\tau)})u(t-\tau) d\tau \\ &= \int_0^t (e^{-(t-\tau)}-e^{-2(t-\tau)}) d\tau, \qquad t\geq 0 \\ &= \left( \frac{1}{2} - e^{-t} + \frac{1}{2}e^{-2t} \right) u(t)
\end{align*}
the key point is that $u(t-\tau)$ as a function of $\tau$ is $1$ for $\tau \leq t$ and zero otherwise (a graph with these transformations - first shift to the right, then invert sign- may help). 
