Smallest palindrome divisible by the input

Haskell, 45 37 34 bytes

(+)>>=until((reverse>>=(==)).show)

Pyth, 7 bytes

*f_I`*Q

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Explanation

*f_I`*QT)Q   implicit endings, Q=input number
 f      )    find the first number T >= 1, which satisfies:
     *QT        product of Q and T
    `           as string
  _I            is invariant under inversion (=palindrom)
*        Q   multiply this number with Q and print

2sable / 05AB1E, 6 / 7 bytes

2sable

[DÂQ#+

Explanation

[         # infinite loop
 D        # duplicate current number
  Â       # bifurcate
   Q#     # if the number is equal to its reverse, break loop
     +    # add input
          # implicitly print

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05AB1E

[DÂQ#¹+

The difference to the 2sable code is that input is only implicit once in 05AB1E, so here we need ¹ to get the first input again.

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Saved 1 byte with 2sable as suggested by Adnan