Evaluate ∫ 1 ∞ ln ⁡ ( 2 x − 1 ) x 2 My approach is to calc ∫ 1 X ln ⁡ ( 2 x − 1 ) x 2 d xand then take the limit for the answer when X → ∞However, I must do something wrong. The correct answer should be 2 ln ⁡ ( 2 ) ∫ 1 X ln ⁡ ( 2 x − 1 ) x 2 d x = [ − 1 x ln ⁡ ( 2 x − 1 ) ] 1 X + ∫ 1 X 1 x × 2 2 x − 1 d x = − 1 X ln ⁡ ( 2 X − 1 ) + 2 ∫ 1 X − 1 x + 2 2 x − 1 d x = − 1 1 X ln ⁡ ( 2 X − 1 ) − 2 ln ⁡ X + 2 ln ⁡ ( 2 X − 1 )Am I wrong? If I'm not, how to proceed?=== EDIT ===After the edit I wonder if this is the correct way to proceed: − 1 X ln ⁡ ( 2 X − 1 ) − 2 ln ⁡ X + 2 ln ⁡ ( 2 X − 1 )The first part will do to zero because of 1 X so we ignore that one, the second and third part: − 2 ln ⁡ X + 2 ln ⁡ ( 2 X − 1 ) = 2 ln ⁡ ( 2 X − 1 X ) = 2 ln ⁡ ( 2 − 1 X ) = 2 ln ⁡ ( 2 )

Question

Evaluate       ∫          1              ∞                  ln      ⁡              (        2        x        −        1        )                    x      2      My approach is to calc      ∫          1              X                  ln      ⁡              (        2        x        −        1        )                    x      2        d  xand then take the limit for the answer when   X  →  ∞However, I must do something wrong. The correct answer should be   2  ln  ⁡  (  2  )      ∫          1              X                  ln      ⁡              (        2        x        −        1        )                    x      2        d  x  =            [      −              1        x            ln      ⁡      (      2      x      −      1      )      ]              1              X        +      ∫          1              X            1    x    ×      2          2      x      −      1        d  x  =  −      1    X    ln  ⁡  (  2  X  −  1  )  +  2      ∫          1              X        −      1    x    +      2          2      x      −      1        d  x  =  −  1      1    X    ln  ⁡  (  2  X  −  1  )  −  2  ln  ⁡  X  +  2  ln  ⁡  (  2  X  −  1  )Am I wrong? If I&#039;m not, how to proceed?=== EDIT ===After the edit I wonder if this is the correct way to proceed:  −      1    X    ln  ⁡  (  2  X  −  1  )  −  2  ln  ⁡  X  +  2  ln  ⁡  (  2  X  −  1  )The first part will do to zero because of       1    X   so we ignore that one, the second and third part:  −  2  ln  ⁡  X  +  2  ln  ⁡  (  2  X  −  1  )  =  2  ln  ⁡      (                  2        X        −        1            X        )    =  2  ln  ⁡      (    2    −          1      X        )    =  2  ln  ⁡  (  2  )

seguidora1e · Accepted Answer

Your derivative is incorrect, it should be      2          2      x      −      1      instead of      1          2      x      −      1      Everything else seems correct.

Haiphongum · Accepted Answer

Another approach :Setting   x  ↦      1    x  , we will obtain      ∫          1              ∞                  ln      ⁡              (        2        x        −        1        )                    x      2           d  x  =      ∫          0              1        ln  ⁡      (                  2        −        x            x        )    d  x  =      ∫          0              1                  [        ln  ⁡  (  2  −  x  )  −  ln  ⁡  x            ]           d  x  .Note that  ∫  ln  ⁡  y     d  y  =  y     ln  ⁡  y  −  y  +  C  ,hence      ∫          1              ∞                  ln      ⁡              (        2        x        −        1        )                    x      2           d  x  =  2  ln  ⁡  2.

int_(1)^(oo) (ln(2x-1))/(x^2) dx Evaluate

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