Having trouble computing $int_3^5frac{t}{1+0.1t} dt $












1












$begingroup$


$$int_3^5frac{t}{1+0.1t} dt $$



For some reason this is equal to:




1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))




I have no idea how to reduce to that.






integration definite-integrals







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$endgroup$












  • $begingroup$
    If you let $x=1+0.1t$, then $t=10x-10$...
    $endgroup$
    – Eleven-Eleven
    Feb 18 at 13:53












  • $begingroup$
    Do you mean $$int_{3}^{5}frac{t}{1+frac{1}{10}t}dt$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Feb 18 at 13:54










  • $begingroup$
    This is not an improper integral, by the way. So, I removed that tag.
    $endgroup$
    – Michael Rybkin
    Feb 18 at 14:29


















1












$begingroup$


$$int_3^5frac{t}{1+0.1t} dt $$



For some reason this is equal to:




1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))




I have no idea how to reduce to that.






integration definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you let $x=1+0.1t$, then $t=10x-10$...
    $endgroup$
    – Eleven-Eleven
    Feb 18 at 13:53












  • $begingroup$
    Do you mean $$int_{3}^{5}frac{t}{1+frac{1}{10}t}dt$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Feb 18 at 13:54










  • $begingroup$
    This is not an improper integral, by the way. So, I removed that tag.
    $endgroup$
    – Michael Rybkin
    Feb 18 at 14:29
















1












1








1





$begingroup$


$$int_3^5frac{t}{1+0.1t} dt $$



For some reason this is equal to:




1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))




I have no idea how to reduce to that.






integration definite-integrals







share|cite|improve this question











$endgroup$




$$int_3^5frac{t}{1+0.1t} dt $$



For some reason this is equal to:




1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))




I have no idea how to reduce to that.






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 20 at 2:20









Michael Rybkin

2,833416




2,833416










asked Feb 18 at 13:51









ximxim

516




516












  • $begingroup$
    If you let $x=1+0.1t$, then $t=10x-10$...
    $endgroup$
    – Eleven-Eleven
    Feb 18 at 13:53












  • $begingroup$
    Do you mean $$int_{3}^{5}frac{t}{1+frac{1}{10}t}dt$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Feb 18 at 13:54










  • $begingroup$
    This is not an improper integral, by the way. So, I removed that tag.
    $endgroup$
    – Michael Rybkin
    Feb 18 at 14:29




















  • $begingroup$
    If you let $x=1+0.1t$, then $t=10x-10$...
    $endgroup$
    – Eleven-Eleven
    Feb 18 at 13:53












  • $begingroup$
    Do you mean $$int_{3}^{5}frac{t}{1+frac{1}{10}t}dt$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Feb 18 at 13:54










  • $begingroup$
    This is not an improper integral, by the way. So, I removed that tag.
    $endgroup$
    – Michael Rybkin
    Feb 18 at 14:29


















$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53






$begingroup$
If you let $x=1+0.1t$, then $t=10x-10$...
$endgroup$
– Eleven-Eleven
Feb 18 at 13:53














$begingroup$
Do you mean $$int_{3}^{5}frac{t}{1+frac{1}{10}t}dt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54




$begingroup$
Do you mean $$int_{3}^{5}frac{t}{1+frac{1}{10}t}dt$$?
$endgroup$
– Dr. Sonnhard Graubner
Feb 18 at 13:54












$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29






$begingroup$
This is not an improper integral, by the way. So, I removed that tag.
$endgroup$
– Michael Rybkin
Feb 18 at 14:29












2 Answers
2






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6












$begingroup$

Hint:



$$frac{t}{1+0.1t} = frac{10cdot(1+0.1t) - 10}{1+0.1t} = 10 - frac{10}{1+0.1t}$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    $$
    frac{x}{1+0.1x}=frac{x}{1+0.1x}cdotfrac{10}{10}=
    frac{10x}{10+x}=10left(frac{x}{10+x}right)=\
    10left(frac{-10+10+x}{10+x}right)=
    10left(frac{-10}{10+x}+frac{10+x}{10+x}right)=
    10left(-frac{10}{10+x}+1right)=\
    10left(1-frac{10}{10+x}right)=10-frac{100}{10+x}.
    $$



    $$
    intleft(10-frac{100}{10+x}right),dx=
    10int,dx-100intfrac{1}{10+x}frac{d}{dx}(10+x),dx=\
    10x-100intfrac{1}{10+x},d(10+x)=
    10x-100ln{|10+x|}+C.
    $$

    $$
    int_3^5frac{t}{1+0.1t},dt=
    bigg[10t-100ln{|10+t|}bigg]_3^5=\
    50-100ln{15}-(30-100ln{13})=
    20-100ln{15}+100ln{13}=\
    20-100(ln{15}-ln{13})=20-100ln{frac{15}{13}}.
    $$

    The answer you gave is equivalent to what I got:
    $$
    frac{1}{0.1}left(2-frac{1}{0.1}left[ln{1.5}-ln{1.3}right]right)=
    10left(2-10left[ln{frac{15}{10}}-ln{frac{13}{10}}right]right)=\
    20-100ln{left(frac{15}{10}divfrac{13}{10}right)}=
    20-100ln{frac{15}{13}}.
    $$




    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Hint:



      $$frac{t}{1+0.1t} = frac{10cdot(1+0.1t) - 10}{1+0.1t} = 10 - frac{10}{1+0.1t}$$






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        Hint:



        $$frac{t}{1+0.1t} = frac{10cdot(1+0.1t) - 10}{1+0.1t} = 10 - frac{10}{1+0.1t}$$






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          Hint:



          $$frac{t}{1+0.1t} = frac{10cdot(1+0.1t) - 10}{1+0.1t} = 10 - frac{10}{1+0.1t}$$






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$frac{t}{1+0.1t} = frac{10cdot(1+0.1t) - 10}{1+0.1t} = 10 - frac{10}{1+0.1t}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 18 at 13:53









          5xum5xum

          91.1k394161




          91.1k394161























              3












              $begingroup$

              $$
              frac{x}{1+0.1x}=frac{x}{1+0.1x}cdotfrac{10}{10}=
              frac{10x}{10+x}=10left(frac{x}{10+x}right)=\
              10left(frac{-10+10+x}{10+x}right)=
              10left(frac{-10}{10+x}+frac{10+x}{10+x}right)=
              10left(-frac{10}{10+x}+1right)=\
              10left(1-frac{10}{10+x}right)=10-frac{100}{10+x}.
              $$



              $$
              intleft(10-frac{100}{10+x}right),dx=
              10int,dx-100intfrac{1}{10+x}frac{d}{dx}(10+x),dx=\
              10x-100intfrac{1}{10+x},d(10+x)=
              10x-100ln{|10+x|}+C.
              $$

              $$
              int_3^5frac{t}{1+0.1t},dt=
              bigg[10t-100ln{|10+t|}bigg]_3^5=\
              50-100ln{15}-(30-100ln{13})=
              20-100ln{15}+100ln{13}=\
              20-100(ln{15}-ln{13})=20-100ln{frac{15}{13}}.
              $$

              The answer you gave is equivalent to what I got:
              $$
              frac{1}{0.1}left(2-frac{1}{0.1}left[ln{1.5}-ln{1.3}right]right)=
              10left(2-10left[ln{frac{15}{10}}-ln{frac{13}{10}}right]right)=\
              20-100ln{left(frac{15}{10}divfrac{13}{10}right)}=
              20-100ln{frac{15}{13}}.
              $$




              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                $$
                frac{x}{1+0.1x}=frac{x}{1+0.1x}cdotfrac{10}{10}=
                frac{10x}{10+x}=10left(frac{x}{10+x}right)=\
                10left(frac{-10+10+x}{10+x}right)=
                10left(frac{-10}{10+x}+frac{10+x}{10+x}right)=
                10left(-frac{10}{10+x}+1right)=\
                10left(1-frac{10}{10+x}right)=10-frac{100}{10+x}.
                $$



                $$
                intleft(10-frac{100}{10+x}right),dx=
                10int,dx-100intfrac{1}{10+x}frac{d}{dx}(10+x),dx=\
                10x-100intfrac{1}{10+x},d(10+x)=
                10x-100ln{|10+x|}+C.
                $$

                $$
                int_3^5frac{t}{1+0.1t},dt=
                bigg[10t-100ln{|10+t|}bigg]_3^5=\
                50-100ln{15}-(30-100ln{13})=
                20-100ln{15}+100ln{13}=\
                20-100(ln{15}-ln{13})=20-100ln{frac{15}{13}}.
                $$

                The answer you gave is equivalent to what I got:
                $$
                frac{1}{0.1}left(2-frac{1}{0.1}left[ln{1.5}-ln{1.3}right]right)=
                10left(2-10left[ln{frac{15}{10}}-ln{frac{13}{10}}right]right)=\
                20-100ln{left(frac{15}{10}divfrac{13}{10}right)}=
                20-100ln{frac{15}{13}}.
                $$




                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $$
                  frac{x}{1+0.1x}=frac{x}{1+0.1x}cdotfrac{10}{10}=
                  frac{10x}{10+x}=10left(frac{x}{10+x}right)=\
                  10left(frac{-10+10+x}{10+x}right)=
                  10left(frac{-10}{10+x}+frac{10+x}{10+x}right)=
                  10left(-frac{10}{10+x}+1right)=\
                  10left(1-frac{10}{10+x}right)=10-frac{100}{10+x}.
                  $$



                  $$
                  intleft(10-frac{100}{10+x}right),dx=
                  10int,dx-100intfrac{1}{10+x}frac{d}{dx}(10+x),dx=\
                  10x-100intfrac{1}{10+x},d(10+x)=
                  10x-100ln{|10+x|}+C.
                  $$

                  $$
                  int_3^5frac{t}{1+0.1t},dt=
                  bigg[10t-100ln{|10+t|}bigg]_3^5=\
                  50-100ln{15}-(30-100ln{13})=
                  20-100ln{15}+100ln{13}=\
                  20-100(ln{15}-ln{13})=20-100ln{frac{15}{13}}.
                  $$

                  The answer you gave is equivalent to what I got:
                  $$
                  frac{1}{0.1}left(2-frac{1}{0.1}left[ln{1.5}-ln{1.3}right]right)=
                  10left(2-10left[ln{frac{15}{10}}-ln{frac{13}{10}}right]right)=\
                  20-100ln{left(frac{15}{10}divfrac{13}{10}right)}=
                  20-100ln{frac{15}{13}}.
                  $$




                  share|cite|improve this answer











                  $endgroup$



                  $$
                  frac{x}{1+0.1x}=frac{x}{1+0.1x}cdotfrac{10}{10}=
                  frac{10x}{10+x}=10left(frac{x}{10+x}right)=\
                  10left(frac{-10+10+x}{10+x}right)=
                  10left(frac{-10}{10+x}+frac{10+x}{10+x}right)=
                  10left(-frac{10}{10+x}+1right)=\
                  10left(1-frac{10}{10+x}right)=10-frac{100}{10+x}.
                  $$



                  $$
                  intleft(10-frac{100}{10+x}right),dx=
                  10int,dx-100intfrac{1}{10+x}frac{d}{dx}(10+x),dx=\
                  10x-100intfrac{1}{10+x},d(10+x)=
                  10x-100ln{|10+x|}+C.
                  $$

                  $$
                  int_3^5frac{t}{1+0.1t},dt=
                  bigg[10t-100ln{|10+t|}bigg]_3^5=\
                  50-100ln{15}-(30-100ln{13})=
                  20-100ln{15}+100ln{13}=\
                  20-100(ln{15}-ln{13})=20-100ln{frac{15}{13}}.
                  $$

                  The answer you gave is equivalent to what I got:
                  $$
                  frac{1}{0.1}left(2-frac{1}{0.1}left[ln{1.5}-ln{1.3}right]right)=
                  10left(2-10left[ln{frac{15}{10}}-ln{frac{13}{10}}right]right)=\
                  20-100ln{left(frac{15}{10}divfrac{13}{10}right)}=
                  20-100ln{frac{15}{13}}.
                  $$





                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 18 at 14:44

























                  answered Feb 18 at 13:59









                  Michael RybkinMichael Rybkin

                  2,833416




                  2,833416






























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