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Dr. belisarius
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The two evaluate to the same values.

f = 1/Sqrt[(x - t)*(y - t)];   In[5]:= v1 = Integrate[f, {t, 0, y}, Assumptions -> {0 < y < x}]   Out[5]=(* Log[(x + y + 2 Sqrt[x y])/(x - y)] *) In[6]:= v2 = Integrate[f, {t, 0, x}, Assumptions -> {0 < x < y}]   Out[6]=(* 2 ArcTanh[Sqrt[x/y]] *) v3 = v2 /. {y -> x, x -> y}   g = v1/v3;   In[23]:= Table[g /. {x -> 10}, {y, 1, 9, 1}] // N   Out[23]=(* {1., 1., 1., 1., 1., 1., 1., 1., 1.} *) 

The two evaluate to the same values.

f = 1/Sqrt[(x - t)*(y - t)];   In[5]:= v1 = Integrate[f, {t, 0, y}, Assumptions -> {0 < y < x}] Out[5]= Log[(x + y + 2 Sqrt[x y])/(x - y)] In[6]:= v2 = Integrate[f, {t, 0, x}, Assumptions -> {0 < x < y}] Out[6]= 2 ArcTanh[Sqrt[x/y]] v3 = v2 /. {y -> x, x -> y}   g = v1/v3;   In[23]:= Table[g /. {x -> 10}, {y, 1, 9, 1}] // N Out[23]= {1., 1., 1., 1., 1., 1., 1., 1., 1.} 

The two evaluate to the same values.

f = 1/Sqrt[(x - t)*(y - t)]; v1 = Integrate[f, {t, 0, y}, Assumptions -> {0 < y < x}]   (* Log[(x + y + 2 Sqrt[x y])/(x - y)] *) v2 = Integrate[f, {t, 0, x}, Assumptions -> {0 < x < y}]   (* 2 ArcTanh[Sqrt[x/y]] *) v3 = v2 /. {y -> x, x -> y} g = v1/v3; Table[g /. {x -> 10}, {y, 1, 9, 1}] // N   (* {1., 1., 1., 1., 1., 1., 1., 1., 1.} *) 
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Asim
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The two evaluate to the same values.

f = 1/Sqrt[(x - t)*(y - t)]; In[5]:= v1 = Integrate[f, {t, 0, y}, Assumptions -> {0 < y < x}] Out[5]= Log[(x + y + 2 Sqrt[x y])/(x - y)] In[6]:= v2 = Integrate[f, {t, 0, x}, Assumptions -> {0 < x < y}] Out[6]= 2 ArcTanh[Sqrt[x/y]] v3 = v2 /. {y -> x, x -> y} g = v1/v3; In[23]:= Table[g /. {x -> 10}, {y, 1, 9, 1}] // N Out[23]= {1., 1., 1., 1., 1., 1., 1., 1., 1.}