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 a 2 
t   cb ln 
a  t 
   4 4  dt .
0 a  t
 a 2 
x   cb ln 
a x   cb ln  x a  x 
Тоді I  4 4 dx    4 4  dx 
0 x  a 0 x  a
  a 2  
x  b2   c ln x  ln  
a   x   a x dx

   4 4   dx  2   cb ln  a  4 4 dx 
0 x  a 0 x  a
a d   b  c ln a x 2 a  b  c ln  a
2
x
 b  c ln  a  4 4  2 arctg 2  2 .
0 a  x a a 0 4a
10.55
 x m arctg x 1 x m arctg x  x m arctg x
I    2 m  1 dx    2 m  1 dx   2 m  1 dx  I  I .
1
2
0 x  1 0 x  1 1 x  1
1 1 m 1
  m 1 1 arctg 1 t arctg
x arctg x 1 t m t dt t
I   dx  x       dt .
2 x 2  1m  1 t 1 t 2 t 2  1m  1
1 0 1 0
t 2  1m 
1 
Далі скористаємось тотожністю arctg x  arctg  при
x 2
x  0. отже,
1 m  1 
m
1 x m arctg x 1 x arctg 1 x arctg x  arctg x 
I   2  m 1  dx   2  m 1  x dx    2  m 1   dx 
0 x 1 0 x 1 0 x 1
1 m 1 m 1  1
 x dx  d x 
    dx  arctg x 
2
2 x 2  1m  1 2  1m  x m 1  1 2  1m 
0 0 0
 2
 .
 8 m   1
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