Antiderivatives and definite integralsCompute the following integral: I = ∫ 0 1 ( ∫ 1 x 1 1 + t 2 d t ) d xMotivation: From time to time, i intend to post here challenging questions -together with proposed answers/solutions- which i frequently give for practise to some of my students (in an introductory calculus course).I am doing this with the understanding that this is not only accepted but furthermore encouraged by the community. The benefits can be twofold: on the one hand this contributes to building a library with well-posed questions and reliable answers (to be used in-class or online) and on the other hand i always hope that this may result in improving (or sometimes correcting) the proposed questions/answers and discovering new approaches/solutions from other users as well.

Question

Antiderivatives and definite integralsCompute the following integral:  I  =      ∫          0              1                  (            ∫          1              x            1          1      +              t        2              d  t            )        d  xMotivation: From time to time, i intend to post here challenging questions -together with proposed answers/solutions- which i frequently give for practise to some of my students (in an introductory calculus course).I am doing this with the understanding that this is not only accepted but furthermore encouraged by the community. The benefits can be twofold: on the one hand this contributes to building a library with well-posed questions and reliable answers (to be used in-class or online) and on the other hand i always hope that this may result in improving (or sometimes correcting) the proposed questions/answers and discovering new approaches/solutions from other users as well.

Maggie Bowman · Accepted Answer

Explanation:                                          ∫            0            1                                ∫            1            x                                              d              t                                      1              +                              t                2                                              ⋅          d          x                                    =                  ∫          0          1                          (          arctan          ⁡          (          x          )          −                      π            4                    )                d        x                                          =                              [                          x        arctan        ⁡        (        x        )        −                  1          2                ln        ⁡        (        1        +                  x          2                )        −                  π          4                x                                            ]                                0          1                =        −                  1          2                ln        ⁡        (        2        )

dream13rxs · Accepted Answer

Step 1Let   f  (  t  )  =      1          1      +              t        2            ,   t  ∈      R   and   F  (  x  )  =      ∫          1              x            1          1      +              t        2              d  t  =      ∫          1              x        f  (  t  )  d  t,   t  ∈      R  Step 2Consequently,       F    ′    (  x  )  =  f  (  x  ),   x  ∈      R   and:  I  =      ∫          0              1                  (            ∫          1              x            1          1      +              t        2              d  t            )        d  x  =      ∫          0              1        F  (  x  )  d  x  =      ∫          0              1        (  x      )    ′    F  (  x  )  d  x  =  =  [  x  F  (  x  )      ]          0              1        −      ∫          0              1        x      F    ′    (  x  )  d  x  =  F  (  1  )  −      ∫          0              1        x  f  (  x  )  d  x  =  0  −      ∫          0              1            x          1      +              x        2              d  x  =  =  −      1    2        ∫          0              1                  (      1      +              x        2                    )        ′                    1      +              x        2              d  x  =  −      1    2    [  ln  ⁡  (  1  +      x    2    )      ]          0              1        =  −            ln      ⁡      2        2

Antiderivatives and definite integrals Compute the following integral: I = <msubsup>

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New Questions in Calculus 1