What is the integral cohomology of $\mathrm{PGL}_n(\mathbb{C})$ as a space? The computation of the cohomology of $\mathrm{GL}_n(\mathbb{C})$ is one of the basic applications of the Serre spectral sequence, using the fiber bundle $\mathrm{GL}_{n-1}(\mathbb{C})\to \mathrm{GL}_n(\mathbb{C}) \to \mathbb{C}^{n} \setminus0$. Using the same idea we can get a fiber bundle $\mathrm{PGL}_{n-1}(\mathbb{C})\to \mathrm{PGL}_n(\mathbb{C}) \to \mathbb{C}P^{n-1}$, but now the base has enough cohomology for the differentials to not be obviously zero.
In fact, I believe the following shows that $H^2(\mathrm{PGL}_n(\mathbb{C}))$ has torsion $\mathbb{Z}/n\mathbb{Z}$ (so not all of the differentials above can be zero), as well as computing $H^*(\mathrm{PGL}_n(\mathbb{C}); \mathbb{Q})$. The important observation is that in the fiber bundle $\mathbb{C}^\times \to \mathrm{GL}_n(\mathbb{C}) \to \mathrm{PGL}_n(\mathbb{C})$, the induced map $\pi_1(\mathbb{C}^\times) \to \pi_1(\mathrm{GL}_n(\mathbb{C}))$ is given by $\mathbb{Z} \xrightarrow{\times n} \mathbb{Z}$. So from the long exact sequence in homotopy, $\pi_1(\mathrm{PGL}_n(\mathbb{C})) = \mathbb{Z}/n\mathbb{Z}$, which shows up in $H^2$.
Further, with $\mathbb{Q}$ coefficients, the map $H^*(\mathrm{GL}_n(\mathbb{C}); \mathbb{Q}) \to H^*(\mathbb{C}^\times; \mathbb{Q})$ is surjective, so by Leray-Hirsch, $H^*(\mathrm{GL}_n(\mathbb{C}); \mathbb{Q}) = H^*(\mathrm{PGL}_n(\mathbb{C}); \mathbb{Q}) \otimes H^*(\mathbb{C}^\times; \mathbb{Q})$, which lets us compute $H^*(\mathrm{PGL}_n(\mathbb{C}); \mathbb{Q})$ – in particular, its Poincaré polynomial is given by $(1+t^3)(1+t^5) \dotsm (1+t^{2n-1})$.
But how does one compute the integral cohomology of $\mathrm{PGL}_n(\mathbb{C})$? 
 A: By arguments outlined in the comments, the projective linear group $PGL_n(\mathbb{C})$ is homotopy equivalent to the projective unitary group $PU(n)$, so it suffices to compute the cohomology of $PU(n)$.  
One way to compute this is to use the Serre spectral sequence for the fiber bundle $U(1) \to U(n) \to PU(n)$.  As the question points out, it's not too hard to use this to deduce rational information.  For mod $p$ cohomology, potentially nontrivial differentials in this spectral sequence make this a more involved calculation.  Nonetheless, this has been worked out in

Baum, Paul F. and Browder, William.
  "The cohomology of quotients of classical groups." 
  Topology 3 (1965), 305–336. [MR0189063]

The idea is that there is another fiber bundle $U(n) \to PU(n) \to \mathbb{C}P^\infty$, obtained by extending the previous fiber bundle once by delooping. The transgressions in this spectral sequence can be analyzed using Chern classes: this principal $U(n)$-bundle is classified by the composite map $\mathbb{C}P^\infty = BU(1) \to BU(1)^{\times n} \to BU(n)$, whose Chern classes, and hence the transgressions, are readily computed.   Working at the prime $p$, the authors find (corollary 4.2): $$H^*(PU(n); \mathbb{Z}/p) \cong \Lambda[x_1, x_2, \ldots, \hat{x_{p^r}}, \ldots, x_n] \otimes P[y]/(y^{p^r}),$$ where $p^r$ is the largest power of $p$ dividing $n$, $\deg x_i = 2i - 1$, and $\deg y = 2$.  
Baum and Browder also provide some integral information in the following corollary (Corollary 4.3).  
