I pentominoes are special, since they must lie in a single row or column, and can only contain the digits 1,2,3,4,7...therefore it must contain each once. Look at the middle column of the center box. The I in the middle square allows only 3 or 4 in this box. Sudoku rules force the square above it to be limited to 3468, and the square below it to 348. In particular, one of these squares must be either 6 or 8. Since a vertically placed I must contain the 7, and cannot start in R9 since R9C5 is already a P, a vertically placed I here must contain all three squares of the center column of the middle square. But one of these is either a 6 or 8, a contradiction. Thus the I must lay horizontally in row 5.
The 9 in row 5 must appear in the middle right square. If it is not in R5C9, then the value in R5C9 is one of 1,2,3,4,7, which is contained in the I; but this forces the I to span R5C5-9, one of whose cells is 9. This is a contradiction, so R5C9 must be 9, forcing R5C4 to be in the I, limiting its values to either 3 or 7. We note this also prevents R5C6 from being 8; if it were 8, the I would have to lie in R5C1-5, forcing the 6 in row 5 to be in the middle right square, a contradiction. The grid thus far, with some letter-based Sudoku restrictions noted:
Looking at the X:
R3C2 cannot be the bottom of the X, for it would leave R1C1 isolated. It cannot be the right of the X, for it would run off the board. If it were the left of the X, then the only way we could place the T would isolate R2C4. Finally, if R3C2 were the center of the X, the remaining cells of the upper left square must be covered with a U, but that would include a 4. So the R3C2 must be the top of the X.
Now consider placements for the T. If the T points left (⊣), then R3C3 is isolated. If the T points up (⊥), then there is an isolated section of 7 cells that cannot be covered with pentominoes. If the T points right (⊢) we isolate 8 cells. If the T points down (T), we isolate 4 cells (leftmost position) or 7 cells (middle position). So the T must point down with its upper left corner in R1C4. The grid thus far:
Some more logic in the upper left:
Notice that there are 10 cells uncovered by pentominoes in the upper left, with only the single egress between the X and T. This means no pentomino can bridge this gap, so there must be two pentominoes that cover these 10 cells. The pentomino containing R3C1 cannot avoid R1, for it would isolate a group of at most 3 cells in this row. And once this pentomino meets row 1, it has to contain a contiguous row of cells from column 1 in row 1, since anything not covered would be isolated. So the only possibilities for this pentomino are T (pointing right), P (in the P orientation), or V (open to the bottom right). T would isolate 2 cells in row 1, and P would forces the other pentomino in this area to be an adjacent P as well, so we must have and V, and thus a P.
Now some Sudoku: since the I in row 5 can start no further left than C3 and end no further right than C8, we can use our known constraints on values in row 5. These constraints force the middle row of the middle square to consist of 3,4,7, allowing us to fix the 6 and 8 in this square. The T in the upper middle square prevents placement of 2, 3 or 8 in those squares, allowing us to restrict the remaining squares to 1, 6, 7, 8 and 9. But SURPRISE! In R3C5, the only possible value is 9, and this forces R2C5 to be 1. Thus R1C4 and R1C6 are 6 and 7 in some order. This lets us force 2,3 in R1C2-3 and 1,9 in R1C7-8. We also find R2C1 is 9. The grid thus far:
The pentomino on R4C4:
The pentomino must extend through R4C4-C6. Since it contains a 9, it can be neither an I, nor an N, and since it contains a 5 it cannot be a V either, forcing it to be a Y or an L, and in particular it must extend to R4C7. But if it is a L, then we must have a pentomino extending from R3C6 -> R2C6 -> R2C7 -> R2C8, which blocks the placement of a P on R1C7. So this must be a Y.
A quick Sudoku hit:
Since R1C2-3 must be 2 and 3, the only possible value for R3C1 is 6...5 and 7 are excluded by the V. This forces the 5 and 8 in the middle left square, which forces R2C3 to be 8 as well. Now, we see R8C3 must also be 5, since the 5 in C2 is in either R2 or R3. This is critical, because:
Place the F:
Looking at the F in R7C2, there are only two possible orientations that do not lead to quick contradictions: either R5C3, R6C2-4 and R7C2, or R6C4, R7C2-4 and R8C3. But the second actually does give us a contradiction, since R8C3 is five, which cannot be in an F. So the F placement is R5C3, R6C2-4 and R7C2. This forces an L to its left R5-8C1 and R8C2. Then the pentomino from R9C2 must be a Z. It must contain R9C3 and R8C3. If it contains R8C4, then adding R8C5 would block placement of the P, while adding R7C4 would isolate R7C3. The grid thus far:
Finishing the pentominous:
Note R8C4-5 must also be in the P. The P cannot extend to R8C6, for it would force a P directly above it. Similarly, it cannot extend to R9C6, as again a P would be forced above, with tail in either R8C6 or R7C7. So the P must extend up to R7C5. Now, the pentomino on R6C5 extends to R6C6. It cannot join the T, since it contains an 8, so it must extend down to R7C6. This leaves only two orientations for the T, pointing down or pointing right, and the pointing down orientation forces 12 uncovered squares needing to be covered by pentominoes. So it must point right, forcing an L and P beneath it. The I extends to R5C8. The shape above the T cannot be a U, since that would leave no place for a 4 in the middle right square, so it must be an L. The rest of the pentominous falls. The grid thus far:
Back to Sudoku:
With a number of blocks in place, we force R2C6 to be 3, forcing the 2 in R3C6 and 8 in R3C4. We can also use the completed I to insert possible values, forcing the 5 in R6C7 and 6 in R5C3. Using the new T, we see the 8 in the lower right square must be in R8C8, forcing an 8 in R4C7. With the F, we see that R6C3 cannot be 4, forcing R6C9 to be 4. The 3/7 pair in R6C1,3 eliminates them as candidates for R4C1-3, and forcing these to be 1,2,4. We now have R3C7=8 and R7C6=8, finishing the 8s. In the upper right, our 4s force R3C8 to be 4, which gives R2C8=5, then R2C9=2 and R2C7=6. With some more basic deductions, we get to:
Continuing in the lower right:
Because R7-8C9 are 1 and 6, R7C8 cannot be 1, and it cannot be 2 or 3 because it contains a T. 9 is also excluded, so we must have a 7 here. This forces the 3s and 7s in the upper right and middle right squares, as well as setting R9C3=7. This leaves only one place for 7 in column 1, causing another cascade of fill-ins. The grid thus far:
Note that the 2 in row 9 has to be in C7 or C8. This forces the 2 in the lower middle box to be in R7C5, forcing R7C1 to be 3. And that moment when you realize you forgot to remove the 3 as a possibility from the T in R8C7...and it falls apart. Finish!