Solve for f ( x ) ∫ 0 x 3 + 1 f ′ ( t ) 1 + f ( t ) d t = ln ⁡ ( x )

Question

Solve for   f  (  x  )      ∫          0                      x        3            +      1                          f        ′            (      t      )              1      +      f      (      t      )        d  t  =  ln  ⁡  (  x  )

Giovani Hickman · Accepted Answer

Notice that                    f        ′            (      t      )              f      (      t      )      +      1        =  [  ln  ∘  (  1  +  f  )      ]    ′    (  t  )  ,so the equation is equivalent to      ∫    0                  x        3            +      1        [  ln  ∘  (  1  +  f  )      ]    ′    (  t  )        d    t  =  ln  ⁡  (  x  )  .By the fundamental theorem of calculus,      ∫    0                  x        3            +      1        [  ln  ∘  (  1  +  f  )      ]    ′    (  t  )        d    t  =  ln  ⁡  [  1  +  f  (      x    3    +  1  )  ]  −  ln  ⁡  [  f  (  0  )  ]  ,so  ln  ⁡  [  1  +  f  (      x    3    +  1  )  ]  −  ln  ⁡  [  1  +  f  (  0  )  ]  =  ln  ⁡      [                  1        +        f        (                  x          3                +        1        )                    1        +        f        (        0        )              ]    =  ln  ⁡  (  x  )  .This is equivalent to            1      +      f      (              x        3            +      1      )              1      +      f      (      0      )        =  x  .This is equivalent to  f  (      x    3    +  1  )  =  [  1  +  f  (  0  )  ]  ⋅      x    −  1.This is equivalent to  f  (      x    3    +  1  )  =  [  1  +  f  (  0  )  ]  ⋅      x    −  1.Let   y  =      x    3    +  1  , hence   x  =            y      −      1        3    . Therefore,  f  (  y  )  =  [  1  +  f  (  0  )  ]  ⋅            y      −      1        3    −  1.Since the equation requires   x  &amp;gt;  0  , it means that   y  &amp;gt;  1. However, this represents a problem, as it means that 0 is not in the domain of f and letting   y  =  0 implies   f  (  0  )  =  −  1  , and we know form the original equation that   f  (  t  )  &amp;gt;  −  1.As such, the original equation has no solutions. If you replace the original integral by      ∫    a                  x        3            +      1        ,where   a  &amp;gt;  1. Accordingly, this results in  f  (  x  )  =  [  1  +  f  (  a  )  ]  ⋅            x      −      1        3    −  1.One can solve for   f  (  a  )  , by having  f  (  a  )  =  [  1  +  f  (  a  )  ]  ⋅            a      −      1        3    −  1  =            a      −      1        3    +            a      −      1        3    ⋅      f    (    a    )    −  1  ,which is equivalent to  [  1  −            a      −      1        3    ]  ⋅      f    (    a    )    =            a      −      1        3    −  1  ,and as long as   a  ≠  2  , this means  f  (  a  )  =                              a          −          1                3            −      1              1      −                        y          −          1                3              =  −  1  ,which means that  f  (  x  )  =  −  1  ,which is not allowed by the original equation. So in fact, the equation can only have solutions if   a  =  2..   f  (  a  ) then becomes a free parameter, and the family  f  (  x  )  =  [  1  +  f  (  2  )  ]  ⋅            x      −      1        3    −  1becomes an infinite family of solutions.

Chaz Blair · Accepted Answer

⇒  3      x    3        f    ′    (      x    3    +  1  )  =  1  +  f  (      x    3    +  1  )    ⟹    x  (  3      x    2    )      f    ′    (      x    3    +  1  )  =  1  +  f  (      x    3    +  1  )    ⟹              3              x        2                    f        ′            (              x        3            +      1      )              1      +      f      (              x        3            +      1      )        =      1    x  Now integrate both sides with respect x and also note that that the derivative of denominator on LHS is written in numerator, hence using substitution and some manipulation , we get  f  (      x    3    +  1  )  =  A  x  −  1where A is a constant and therefore                              f          (          x          )          =          A          (          x          −          1                      )                                          1                3                                              −          1

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