Fourier transform of an integrator filter
As you know, we don't solve homework problems here, but you can get some hints. I don't know if you're really required to solve the problem in the frequency domain, but I would suggest to compute the output of the integrator in the time domain:

$$y(t)=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau=\int_{-\infty}^{\infty}e^{-\tau/T}u(\tau)u(t-\tau)d\tau\tag{1}$$

This integral turns into something very simple after noting how the infinite integration limits change to finite values when taking into account the two step functions. I'm sure you can continue from here.

If you really want to know the Fourier transform of the unit step $u(t)$, it can be found in most text books on signal processing, and in most Fourier transform tables:

$$\mathcal{F}\{u(t)\}=U(f)=\frac12\delta(f)+\frac{1}{j2\pi f}\tag{2}$$

It's not straightforward to derive this result yourself, so I don't think that this was expected of you.

EDIT:

If you solve the problem in the frequency domain, then with $(2)$ and with $X(f)=\frac{T}{1+j2\pi fT}$ you obtain

$$\begin{align}Y(f)&amp;=X(f)U(f)\\&amp;=\frac{T}{1+j2\pi fT}\left[\frac12\delta(f)+\frac{1}{j2\pi f}\right]\\&amp;=\frac{T}{2}\delta(f)+\frac{T}{1+j2\pi fT}\cdot\frac{1}{j2\pi f}\tag{3}\end{align}$$

where the last equality follows from the fact that $X(f)\delta(f)=X(0)\delta(f)$ is $X(f)$ is continuous at $f=0$.