y(t) a of an integrator circuit
The input and the integrator output are shown below:



So, you should consider two cases: $t&lt;0$ and $t&gt;0$. 

In the first case ($t&lt;0$), you have 
$$\begin{align}
y(t) &amp;= \int_{-\infty}^t x(\tau)d\tau = -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}u(-\tau)d\tau \\
&amp;= -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}d\tau = -\frac{1}{T}Te^{\frac{\tau}{T}}\Big]_{-\infty}^t \\
&amp;= -e^{\frac{\tau}{T}}\Big]_{-\infty}^{t} = -(e^{\frac{t}{T}} - 0) = -e^{\frac{t}{T}}
\end{align}$$
again, for $t&lt;0$. So now you have
$$y(t) = \left\{\begin{array}{ll}
-e^{\frac{t}{T}}, &amp; t &lt; 0 \\
? \quad \:\:\:, &amp; t &gt; 0
\end{array}\right.$$
and you're looking for the other branch of the output. Can you show that it is $$y(t) = -e^{-\frac{t}{T}}, \quad t &gt; 0?$$

If you do, then you'll get
$$y(t) = \left\{\begin{array}{ll}
-e^{\frac{t}{T}}, &amp; t &lt; 0 \\
-e^{-\frac{t}{T}},  &amp; t &gt; 0
\end{array}\right.= -e^{-\frac{|t|}{T}}$$ for all $t$.