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Verified answer
InTegraL Tentu∫dx /x² = 1/2 [2a...-1]
∫x^-2 dx = 1/2
-1/x [2a...-1] = 1/2
-1/(2a) - (-1/-1) = 1/2
-1/(2a) - 1 = 1/2
-1/(2a) = 1 + 1/2
-1/(2a) = 3/2
-1/a = 3
a = -1/3