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My answer:

First steps:

Some initial non-shaded deductions to force connectivity and to prevent any groups of more than 4 tiles from being shaded

That forces some shaded squares to extend out in order to have enough space for tetrominoes; the upper left one is forced to be a T so as to not trap an unshaded square in the corner

A tricky deduction:

The orange squares must be part of an L tetromino (they can't make anything else legally). If the blue square is unshaded, that forces the green squares to be shaded to avoid making an illegal 3rd T. However, now R2C7 can only make Ls and Ts, which are all used up elsewhere. So the blue square must be shaded.

Working on the right side:

We can now set several squares around the newly-made 3-block to unshaded to avoid another L

The newly-made 3-block must be the final T, so the other 3-block must be an S

Another tricky deduction

If R3C8 is shaded, that forces all the green squares to be shaded and the blue square to be unshaded. Now the orange square can only form Ls, Ts, and Ss, all of which are used up (the 3-block to its immediate right must be an S). Therefore R3C8 must be unshaded

Working on the middle:

Basic extensions from unshading R3C8. Note that the orange 2-block must be an S (no Ls or Ts are left)

The 2-block that was part of the i cannot be an S (none are left), so it must connect with R2C9 above it to form an I

Using up the remaining shapes:

If the orange S has its second part to the left, it will isolate some unshaded cells. Therefore its second part is to its right.

R5C11 can't be a T, L, or S so it must be an O

The right side must have an O and an I, and there is only one way to fit them

The final L must be turned to the left so as to not isolate the unshaded squares in the upper right, and then we can set all remaining squares to unshaded

My answer:

First steps:

Some initial non-shaded deductions to force connectivity and to prevent any groups of more than 4 tiles from being shaded

That forces some shaded squares to extend out in order to have enough space for tetrominoes; the upper left one is forced to be a T so as to not trap an unshaded square in the corner

A tricky deduction:

The orange squares must be part of an L tetromino (they can't make anything else legally). If the blue square is unshaded, that forces the green squares to be shaded to avoid making an illegal 3rd L. However, now R2C7 can only make Ls and Ts, which are all used up elsewhere. So the blue square must be shaded.

Working on the right side:

We can now set several squares around the newly-made 3-block to unshaded to avoid another L

The newly-made 3-block must be the final T, so the other 3-block must be an S

Another tricky deduction

If R3C8 is shaded, that forces all the green squares to be shaded and the blue square to be unshaded. Now the orange square can only form Ls, Ts, and Ss, all of which are used up (the 3-block to its immediate right must be an S). Therefore R3C8 must be unshaded

Working on the middle:

Basic extensions from unshading R3C8. Note that the orange 2-block must be an S (no Ls or Ts are left)

The 2-block that was part of the i cannot be an S (none are left), so it must connect with R2C9 above it to form an I

Using up the remaining shapes:

If the orange S has its second part to the left, it will isolate some unshaded cells. Therefore its second part is to its right.

R5C11 can't be a T, L, or S so it must be an O

The right side must have an O and an I, and there is only one way to fit them

The final L must be turned to the left so as to not isolate the unshaded squares in the upper right, and then we can set all remaining squares to unshaded

My answer:

First steps:

Some initial non-shaded deductions to force connectivity and to prevent any groups of more than 4 tiles from being shaded

That forces some shaded squares to extend out in order to have enough space for tetrominoes; the upper left one is forced to be a T so as to not trap an unshaded square in the corner

An invalid deduction:

My thought process was that if the blue square was unshaded, then the only way to place 2 Is would also force too many Ls (green is forced-shaded). I overlooked how an I could go on the bottom row near the middle. Any suggestions for a better deduction are welcome.

Using that invalid deduction:

The new I tetromino forces quite a bit of shading & unshading. Note that now the blue squares must be part of the other L tetromino, and the green square part of the other T tetromino

The tetromino in the middle of the left can't be a T, so it must be an S

If the shaded square in R3C6 doesn't make a tetromino with R2C7, then it will make an L tetromino. But it can't make an L tetromino - that piece was already placed elsewhere - so it must make a T tetromino with R2C7.

Using up the remaining shapes:

R5C6 can't be a T or an L, so it must be an S

R5C11 can't be a T, L, or S so it must be an O

The right side must have an O and an I, and there is only one way to fit them

The final L must be turned to the left so as to not isolate the unshaded squares in the upper right, and then we can set all remaining squares to unshaded

My answer:

First steps:

Some initial non-shaded deductions to force connectivity and to prevent any groups of more than 4 tiles from being shaded

That forces some shaded squares to extend out in order to have enough space for tetrominoes; the upper left one is forced to be a T so as to not trap an unshaded square in the corner

A tricky deduction:

The orange squares must be part of an L tetromino (they can't make anything else legally). If the blue square is unshaded, that forces the green squares to be shaded to avoid making an illegal 3rd T. However, now R2C7 can only make Ls and Ts, which are all used up elsewhere. So the blue square must be shaded.

Working on the right side:

We can now set several squares around the newly-made 3-block to unshaded to avoid another L

The newly-made 3-block must be the final T, so the other 3-block must be an S

Another tricky deduction

If R3C8 is shaded, that forces all the green squares to be shaded and the blue square to be unshaded. Now the orange square can only form Ls, Ts, and Ss, all of which are used up (the 3-block to its immediate right must be an S). Therefore R3C8 must be unshaded

Working on the middle:

Basic extensions from unshading R3C8. Note that the orange 2-block must be an S (no Ls or Ts are left)

The 2-block that was part of the i cannot be an S (none are left), so it must connect with R2C9 above it to form an I

Using up the remaining shapes:

If the orange S has its second part to the left, it will isolate some unshaded cells. Therefore its second part is to its right.

R5C11 can't be a T, L, or S so it must be an O

The right side must have an O and an I, and there is only one way to fit them

The final L must be turned to the left so as to not isolate the unshaded squares in the upper right, and then we can set all remaining squares to unshaded

My answer:

Explanation to come!

My answer:

First steps:

Some initial non-shaded deductions to force connectivity and to prevent any groups of more than 4 tiles from being shaded

That forces some shaded squares to extend out in order to have enough space for tetrominoes; the upper left one is forced to be a T so as to not trap an unshaded square in the corner

An invalid deduction:

My thought process was that if the blue square was unshaded, then the only way to place 2 Is would also force too many Ls (green is forced-shaded). I overlooked how an I could go on the bottom row near the middle. Any suggestions for a better deduction are welcome.

Using that invalid deduction:

The new I tetromino forces quite a bit of shading & unshading. Note that now the blue squares must be part of the other L tetromino, and the green square part of the other T tetromino

The tetromino in the middle of the left can't be a T, so it must be an S

If the shaded square in R3C6 doesn't make a tetromino with R2C7, then it will make an L tetromino. But it can't make an L tetromino - that piece was already placed elsewhere - so it must make a T tetromino with R2C7.

Using up the remaining shapes:

R5C6 can't be a T or an L, so it must be an S

R5C11 can't be a T, L, or S so it must be an O

The right side must have an O and an I, and there is only one way to fit them

The final L must be turned to the left so as to not isolate the unshaded squares in the upper right, and then we can set all remaining squares to unshaded