The function is ∫ 1 1 + 2 x d xUsing u = 1 + 2 x , d u = 1 2 x d x. Which gives us 2 x ∫ 1 u d uThe final answer of this would give 2 x ln ⁡ | 1 + 2 x | , but when I checked my answer the correct answer is 2 x − ln ⁡ | 1 + 2 x |

Question

The function is  ∫      1          1      +              2        x              d  xUsing   u  =  1  +      2    x  ,   d  u  =      1          2      x        d  x. Which gives us      2    x    ∫      1    u    d  uThe final answer of this would give       2    x    ln  ⁡      |    1  +      2    x        |  , but when I checked my answer the correct answer is       2    x    −  ln  ⁡      |    1  +      2    x        |

Milton Gilmore · Accepted Answer

The way you have solved the problem is wrong. When you make the substitution  u  =  1  +      2    x      ⟹    x  =            (      u      −      1              )        2              2      ⟹    d  x  =  (  u  −  1  )  d  uTherefore the integral then is:  ∫            u      −      1        u    d  u  =  u  −  log  ⁡  (      |    u      |    )  +  C  =  1  +      2    x    −  log  ⁡  (      |    1  +      2    x        |    )  +  C  =      2    x    −  log  ⁡  (      |    1  +      2    x        |    )  +      C    ′

dannigurl21ck2 · Accepted Answer

Let   u  =  1  +      2    x  , then  x  =      1    2    (  u  −  1      )    2  and  d  x  =  (  u  −  1  )  d  uTherefore                    ∫                  1                      1            +                          2              x                                      d        x                            =        ∫                  1          u                (        u        −        1        )        d        u                                          =        ∫                  (          1          −                      1            u                    )                d        u                                          =        u        −        log        ⁡                  |                u                  |                +        c                                          =        (        1        +                  2          x                )        −        log        ⁡        (        1        +                  2          x                )        +        c

Integration of a natural logarithmic function using substitution int (1)/(1+sqrt(2x))dx

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New Questions in Calculus 1