Python, 46 bytes

f=lambda s:s*(s==s[::-1])or s[0]+f(s[1:])+s[0]

If the string is a palindrome, return it. Otherwise, sandwich the first letter around the recursive result for the remainder of the string.

Example breakdown:

f(bonobo)
b  f(onobo) b
b o f(nobo) o b 
b o n f(obo) n o b
b o n obo n o b

Pyth, 11 bytes

.VkI_IJ+zbB

Test suite

This code exploits a bug in the current version of Pyth, commit b93a874. The bug is that _IJ+zb is parsed as if it was q_J+zbJ+zb, which is equivalent to _I+zb+zb, when it should (by the design intention of Pyth) be parsed as q_J+zbJ, which is equivalent to _I+zb. This allows me to save a byte - after the bug is fixed, the correct code will be .VkI_IJ+zbJB. I'll explain that code instead.

Basically, the code brute forces over all possible strings until it finds the shortest string that can be appended to the input to form a palindrome, and outputs the combined string.

.VkI_IJ+zbJB
                z = input()
.Vk             For b in possible strings ordered by length,
       +zb      Add z and b,
      J         Store it in J,
    _I          Check if the result is a palindrome,
   I            If so,
          J     Print J (This line doesn't actually exist, gets added by the bug.
          B     Break.

Pyth - 16 12 bytes

4 bytes saved thanks to @FryAmTheEggman.

FGITW, much golfing possible.

h_I#+R_Q+k._

Test Suite.

Haskell, 36 bytes

f s|s==reverse s=s|h:t<-s=h:f t++[h]

More readably:

f s
 |s==reverse s = s
 |(h:t)<-s     = h:(f t)++[h]

If the string is a palindrome, return it. Otherwise, sandwich the first letter around the recursive result for the tail of the string.

The string s is split into h:t in the second guard, obviating a filler 1>0 for this case. This is shorter than doing s@(h:t) for the input.

Retina, 29 25

$
¶$_
O^#r`.\G
(.+)¶\1
$1

Try it online!

Big thanks to Martin for 11 bytes saved!

This just creates a reversed copy of the string and smooshes them together. The only really fancy part of this is the reversing method: O^#r`.\G, which is done by using sort mode. We sort the letters of the second string (the ones that aren't newlines and are consecutive from the end of the string, thanks to the \G) by their numeric value, which, since there are no numbers, is 0. Then we reverse the order of the results of this stable sort with the ^ option. All credit for the fancy use of \G belongs to Martin :)

CJam, 18

q__,,{1$>_W%=}#<W%

Try it online

Explanation:

q         read the input
__        make 2 copies
,,        convert the last one to a range [0 … length-1]
{…}#      find the first index that satisfies the condition:
  1$>     copy the input string and take the suffix from that position
  _W%=    duplicate, reverse and compare (palindrome check)
<         take the prefix before the found index
W%        reverse that prefix
          at the end, the stack contains the input string and that reversed prefix

JavaScript (ES6), 92 bytes

(s,a=[...s],r=a.reverse().join``)=>s.slice(0,a.findIndex((_,i)=>r.startsWith(s.slice(i))))+r

Computes and slices away the overlap between the original string and its reversal.

Octave, 78 bytes

function p=L(s) d=0;while ~all(diag(s==rot90(s),d++)) p=[s fliplr(s(1:d))];end

ideone still fails for named functions, but here is a test run of the code as a program.

Jelly, 11 10 bytes

ṫỤfU$Ḣœ^;U

Try it online!

How it works

ṫỤfU$Ḣœ^;U  Main link. Argument: s (string)

 Ụ          Yield all indices of s, sorted by their corr. character values.
ṫ           Tail; for each index n, remove all characters before thr nth.
            This yields the list of suffixes of s, sorted by their first character,
            then (in descending order) by length.
    $       Combine the two links to the left into a chain:
   U        Upend; reverse all suffixes.
  f         Filter; only keep suffixes that are also reversed suffixes.
            This gives the list of all palindromic suffixes. Since all of them
            start with the same letter, they are sorted by length.
     Ḣ      Head; select the first, longest palindromic suffix.
      œ^    Multiset symmetric difference; chop the selected suffix from s.
         U  Upend; yield s, reversed.
        ;   Concatenate the results to the left and to the right.

Lua, 89 Bytes

I beat the Javascript! \o/

It's a full program, takes its input as command-line argument.

i=1s=...r=s:reverse()while(s:sub(i)~=r:sub(0,#r-i+1))do i=i+1 end print(s..r:sub(#r-i+2))

ungolfed

i=1                             -- initialise i at 1 as string are 1-indexed in lua
s=...                           -- use s as a shorthand for the first argument
r=s:reverse()                   -- reverse the string s and save it into r
while(s:sub(i)~=r:sub(0,#r-i+1))-- iterate while the last i characters of s
do                              -- aren't equals to the first i characters of r
  i=i+1                         -- increment the number of character to skip
end
print(s..r:sub(#r-i+2))         -- output the merged string

05AB1E, 18 bytes

Code:

©gF¹ðN×®«ðñD®åi,q

Uses CP-1252 encoding. Try it online!

Pyke, 15 bytes

D_Dlhm>m+#D_q)e

Try it here!

J, 20 bytes

,[|.@{.~(-:|.)\.i.1:

This is a monadic verb. Try it here. Usage:

   f =: ,[|.@{.~(-:|.)\.i.1:
   f 'race'
'racecar'

Explanation

I'm using the fact that the palindromization of S is S + reverse(P), where P is the shortest prefix of S whose removal results in a palindrome. In J, it's a little clunky to do a search for the first element of an array that satisfies a predicate; hence the indexing.

,[|.@{.~(-:|.)\.i.1:  Input is S.
        (    )\.      Map over suffixes of S:
         -:             Does it match
           |.           its reversal? This gives 1 for palindromic suffixes and 0 for others.
                i.1:  Take the first (0-based) index of 1 in that array.
 [   {.~              Take a prefix of S of that length: this is P.
  |.@                 Reverse of P.
,                     Concatenate it to S.

Haskell, 68 bytes

import Data.List
f i=[i++r x|x<-inits i,i++r x==x++r i]!!0
r=reverse

Usage example: f "abcb" -> "abcba".

Search through the inits of the input i (e.g. inits "abcb" -> ["", "a", "ab", "abc", "abcb"]) until you find one where it's reverse appended to i builds a palindrome.

MATL, 17 16 bytes

Loosely inspired in @aditsu's CJam answer.

`xGt@q:)PhttP=A~

Try it online!

Explanation

`        % Do...while loop
  x      %   Delete top of stack, which contains a not useful result from the
         %   iteration. Takes input implicitly on first iteration, and deletes it
  G      %   Push input
  t      %   Duplicate
  @q:    %   Generate range [1,...,n-1], where n is iteration index. On the  first
         %   iteration this is an empty array
  )      %   Use that as index into copy of input string: get its first n elements
  Ph     %   Flip and concatenate to input string
  t      %   Duplicate. This will be the final result, or will be deleted at the
         %   beginning of next iteration
  tP     %   Duplicate and flip
  =A~    %   Compare element-wise. Is there some element different? If so, the
         %   result is true. This is the loop condition, so go there will be a 
         %   new iteration. Else the loop is exited with the stack containing
         %   the contatenated string
         % End loop implicitly
         % Display stack contents implicitly

Ruby, 44 bytes

This answer is based on xnor's Python and Haskell solutions.

f=->s{s.reverse==s ?s:s[0]+f[s[1..-1]]+s[0]}

Perl, 37 bytes

Based on xnor's answer.

Includes +2 for -lp

Run with input on STDIN, e.g.

palindromize.pl <<< bonobo

palindromize.pl:

#!/usr/bin/perl -lp
s/.//,do$0,$_=$&.$_.$&if$_!~reverse

Brachylog, 16 bytes

:AcCrC,[C]:"~s"w

This could have been 6 bytes if that language sucked a bit less…

This expects a character codes string as input and no output, e.g. brachylog_main(`bonobo`,_).

Explanation

:AcCrC              Unify C with a codes string which is a palindrome and which is the result
                    of the concatenation of the input with another unknown string A

      ,[C]:"~s"w    This is only here to print C as a codes string. This wouldn't be needed
                    if concatenation worked properly (i.e. with one of the two inputs not 
                    unified) on strings.

Oracle SQL 11.2, 195 bytes

SELECT MIN(p)KEEP(DENSE_RANK FIRST ORDER BY LENGTH(p))FROM(SELECT:1||SUBSTR(REVERSE(:1),LEVEL+1)p FROM DUAL WHERE SUBSTR(:1,-LEVEL,LEVEL)=SUBSTR(REVERSE(:1),1,LEVEL)CONNECT BY LEVEL<=LENGTH(:1));

Un-golfed

SELECT MIN(p)KEEP(DENSE_RANK FIRST ORDER BY LENGTH(p))
FROM (
       SELECT :1||SUBSTR(REVERSE(:1),LEVEL+1)p 
       FROM   DUAL 
       WHERE  SUBSTR(:1,-LEVEL,LEVEL)=SUBSTR(REVERSE(:1),1,LEVEL)
       CONNECT BY LEVEL<=LENGTH(:1)
     );

Seriously, 34 bytes

╩╜lur`╜╨"Σ╜+;;R=*"£M`MΣ;░p╜;;R=I.

The last character is a non-breaking space (ASCII 127 or 0x7F).

Try it online!

Explanation:

╩╜lur`╜╨"Σ╜+;;R=*"£M`MΣ;░p╜;;R=I.<NBSP>
╩                                        push inputs to registers (call the value in register 0 "s" for this explanation)
 ╜lur                                    push range(0, len(s)+1)
     `              `M                   map (for i in a):
      ╜╨                                   push all i-length permutations of s
        "        "£M                       map (for k in perms):
         Σ╜+                                 push s+''.join(k) (call it p)
            ;;R=                             palindrome test
                *                            multiply (push p if palindrome else '')
                      Σ                  summation (flatten lists into list of strings)
                       ;░                filter truthy values
                         p               pop first element (guaranteed to be shortest, call it x)
                          ╜;;R=I         pop x, push s if s is palindromic else x
                                .<NBSP>  print and quit

C#, 202 bytes

I tried.

class P{static void Main(string[]a){string s=Console.ReadLine(),o=new string(s.Reverse().ToArray()),w=s;for(int i=0;w!=new string(w.Reverse().ToArray());){w=s.Substring(0,i++)+o;}Console.WriteLine(w);}}

Ungolfed:

class P
{
    static void Main(string[] a)
    {
        string s = Console.ReadLine(), o = new string(s.Reverse().ToArray()), w = s;
        for(int i = 0; w!=new string(w.Reverse().ToArray());)
        {
            w = s.Substring(0, i++) + o;
        }
        Console.WriteLine(w);
        Console.ReadKey();
    }

}

Can anyone provide me with any ideas to group the two calls to .Reverse().ToArray() ? A separate method is more bytes.