Find the $frac{d}{dx}$ of $g(x)=int_{2x}^{3x} frac{u^2-1}{u^2+1}du$












0












$begingroup$


First I split the integral



$$int_{2x}^{0} f(u)du+int_{0}^{3x}f(u)du$$
Then differentiate with respect to x
$$frac{d}{du}left[int_{2x}^{0}f(u)du+ int_{0}^{3x}f(u)duright]frac{dx}{du}$$
In my notes after diff. with respect to x I have
$$f'(x)=frac{4x}{(x^2+1)^2}$$
$$f'(2x)=frac{8x}{(2x^2+1)^2}$$
$$f'(3x)=frac{12x}{(3x^2+1)^2}$$
Then for an answer I get
$$g'(x)=frac{-2(4x^2-1)}{4x^2+1}+frac{3(9x^2-1}{9x^2+1}$$



And I can't understand how I arrived at that answer.






calculus







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  • $begingroup$
    The fourth line does not make sense: $u$ is nothing but a dummy variable.
    $endgroup$
    – user587192
    Dec 10 '18 at 4:52
















0












$begingroup$


First I split the integral



$$int_{2x}^{0} f(u)du+int_{0}^{3x}f(u)du$$
Then differentiate with respect to x
$$frac{d}{du}left[int_{2x}^{0}f(u)du+ int_{0}^{3x}f(u)duright]frac{dx}{du}$$
In my notes after diff. with respect to x I have
$$f'(x)=frac{4x}{(x^2+1)^2}$$
$$f'(2x)=frac{8x}{(2x^2+1)^2}$$
$$f'(3x)=frac{12x}{(3x^2+1)^2}$$
Then for an answer I get
$$g'(x)=frac{-2(4x^2-1)}{4x^2+1}+frac{3(9x^2-1}{9x^2+1}$$



And I can't understand how I arrived at that answer.






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    The fourth line does not make sense: $u$ is nothing but a dummy variable.
    $endgroup$
    – user587192
    Dec 10 '18 at 4:52














0












0








0





$begingroup$


First I split the integral



$$int_{2x}^{0} f(u)du+int_{0}^{3x}f(u)du$$
Then differentiate with respect to x
$$frac{d}{du}left[int_{2x}^{0}f(u)du+ int_{0}^{3x}f(u)duright]frac{dx}{du}$$
In my notes after diff. with respect to x I have
$$f'(x)=frac{4x}{(x^2+1)^2}$$
$$f'(2x)=frac{8x}{(2x^2+1)^2}$$
$$f'(3x)=frac{12x}{(3x^2+1)^2}$$
Then for an answer I get
$$g'(x)=frac{-2(4x^2-1)}{4x^2+1}+frac{3(9x^2-1}{9x^2+1}$$



And I can't understand how I arrived at that answer.






calculus







share|cite|improve this question









$endgroup$




First I split the integral



$$int_{2x}^{0} f(u)du+int_{0}^{3x}f(u)du$$
Then differentiate with respect to x
$$frac{d}{du}left[int_{2x}^{0}f(u)du+ int_{0}^{3x}f(u)duright]frac{dx}{du}$$
In my notes after diff. with respect to x I have
$$f'(x)=frac{4x}{(x^2+1)^2}$$
$$f'(2x)=frac{8x}{(2x^2+1)^2}$$
$$f'(3x)=frac{12x}{(3x^2+1)^2}$$
Then for an answer I get
$$g'(x)=frac{-2(4x^2-1)}{4x^2+1}+frac{3(9x^2-1}{9x^2+1}$$



And I can't understand how I arrived at that answer.






calculus




calculus






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asked Dec 10 '18 at 3:08









Eric BrownEric Brown

737




737












  • $begingroup$
    The fourth line does not make sense: $u$ is nothing but a dummy variable.
    $endgroup$
    – user587192
    Dec 10 '18 at 4:52


















  • $begingroup$
    The fourth line does not make sense: $u$ is nothing but a dummy variable.
    $endgroup$
    – user587192
    Dec 10 '18 at 4:52
















$begingroup$
The fourth line does not make sense: $u$ is nothing but a dummy variable.
$endgroup$
– user587192
Dec 10 '18 at 4:52




$begingroup$
The fourth line does not make sense: $u$ is nothing but a dummy variable.
$endgroup$
– user587192
Dec 10 '18 at 4:52










3 Answers
3






active

oldest

votes


















0












$begingroup$

Let $F(u)$ be a function such that $F'(u)=f(u)$ with
$$
f(u)=frac{u^2-1}{u^2+1}.
$$
The fundamental theorem of calculus tells you that
$$
g(x)=F(3x)-F(2x).
$$
Now by the chain rule you get
$$
g'(x)=3cdot F'(3x)-2F'(2x)=3f(3x)-2f(2x)=frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}.
$$




Alternatively, you could apply directly the Leibniz integral rule.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So after plugging in I get $frac{27x^2-3}{27x^2+3}-frac{8x^2-2}{8x^2+2}$ Then how do i arrive at the answer?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:53












  • $begingroup$
    You did the wrong "plugging in". $3f(3x)-2f(2x)=3cdotfrac{(3x)^2-1}{(3x^2)^2+1}-2cdotfrac{(2x)^2-1}{(2x^2)^2+1}={color{red}{frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}}}$ is your answer.
    $endgroup$
    – user587192
    Dec 10 '18 at 13:25





















1












$begingroup$

Hint



$$frac{d}{dx}int_{a(x)}^{b(x)}f(u) , du=f(b(x)) frac{d}{dx}(b(x))-f(a(x)) frac{d}{dx}(a(x))$$
In your case $a(x)=2x, b(x)=3x$ and $f(u)=frac{u^2-1}{u^2+1}$. You don't need the derivatives of $f$ that you have computed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so I sub in 3x for u to get $3x^2$?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 3:44










  • $begingroup$
    @EricBrown Actually $f(3x)=frac{(3x)^2-1}{(3x)^2+1}$.
    $endgroup$
    – Anurag A
    Dec 10 '18 at 3:48












  • $begingroup$
    so after plugging in I get $frac{9x^2-1}{9x^2+1}(3)-frac{4x^2-1}{4x^2+1}(2)$. Then I multiply the two fractions to get $$frac{27x^2-3}{27x^3+3}-frac{8x^2-2}{8x^2+2}$$
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:00












  • $begingroup$
    @EricBrown Yes that's correct.
    $endgroup$
    – Anurag A
    Dec 10 '18 at 4:02










  • $begingroup$
    Do I still need to integrate the two fractions?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:35



















1












$begingroup$

$$I(u)=intfrac{u^2-1}{u^2+1}mathrm du$$
$$I(u)=intfrac{u^2+1}{u^2+1}mathrm du-2intfrac{mathrm du}{u^2+1}$$
$$I(u)=u-2intfrac{mathrm du}{u^2+1}$$
Using the substitution $tan t=u$, we arrive at
$$I(u)=u-2arctan u$$
And from $g(x)=I(3x)-I(2x)$,
$$g(x)=x+arctan2x-arctan3x$$
Hence
$$g'(x)=1+frac2{1+4x^2}-frac3{1+9x^2}$$






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Let $F(u)$ be a function such that $F'(u)=f(u)$ with
    $$
    f(u)=frac{u^2-1}{u^2+1}.
    $$
    The fundamental theorem of calculus tells you that
    $$
    g(x)=F(3x)-F(2x).
    $$
    Now by the chain rule you get
    $$
    g'(x)=3cdot F'(3x)-2F'(2x)=3f(3x)-2f(2x)=frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}.
    $$




    Alternatively, you could apply directly the Leibniz integral rule.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So after plugging in I get $frac{27x^2-3}{27x^2+3}-frac{8x^2-2}{8x^2+2}$ Then how do i arrive at the answer?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:53












    • $begingroup$
      You did the wrong "plugging in". $3f(3x)-2f(2x)=3cdotfrac{(3x)^2-1}{(3x^2)^2+1}-2cdotfrac{(2x)^2-1}{(2x^2)^2+1}={color{red}{frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}}}$ is your answer.
      $endgroup$
      – user587192
      Dec 10 '18 at 13:25


















    0












    $begingroup$

    Let $F(u)$ be a function such that $F'(u)=f(u)$ with
    $$
    f(u)=frac{u^2-1}{u^2+1}.
    $$
    The fundamental theorem of calculus tells you that
    $$
    g(x)=F(3x)-F(2x).
    $$
    Now by the chain rule you get
    $$
    g'(x)=3cdot F'(3x)-2F'(2x)=3f(3x)-2f(2x)=frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}.
    $$




    Alternatively, you could apply directly the Leibniz integral rule.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So after plugging in I get $frac{27x^2-3}{27x^2+3}-frac{8x^2-2}{8x^2+2}$ Then how do i arrive at the answer?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:53












    • $begingroup$
      You did the wrong "plugging in". $3f(3x)-2f(2x)=3cdotfrac{(3x)^2-1}{(3x^2)^2+1}-2cdotfrac{(2x)^2-1}{(2x^2)^2+1}={color{red}{frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}}}$ is your answer.
      $endgroup$
      – user587192
      Dec 10 '18 at 13:25
















    0












    0








    0





    $begingroup$

    Let $F(u)$ be a function such that $F'(u)=f(u)$ with
    $$
    f(u)=frac{u^2-1}{u^2+1}.
    $$
    The fundamental theorem of calculus tells you that
    $$
    g(x)=F(3x)-F(2x).
    $$
    Now by the chain rule you get
    $$
    g'(x)=3cdot F'(3x)-2F'(2x)=3f(3x)-2f(2x)=frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}.
    $$




    Alternatively, you could apply directly the Leibniz integral rule.






    share|cite|improve this answer









    $endgroup$



    Let $F(u)$ be a function such that $F'(u)=f(u)$ with
    $$
    f(u)=frac{u^2-1}{u^2+1}.
    $$
    The fundamental theorem of calculus tells you that
    $$
    g(x)=F(3x)-F(2x).
    $$
    Now by the chain rule you get
    $$
    g'(x)=3cdot F'(3x)-2F'(2x)=3f(3x)-2f(2x)=frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}.
    $$




    Alternatively, you could apply directly the Leibniz integral rule.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 4:48









    user587192user587192

    1,868315




    1,868315












    • $begingroup$
      So after plugging in I get $frac{27x^2-3}{27x^2+3}-frac{8x^2-2}{8x^2+2}$ Then how do i arrive at the answer?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:53












    • $begingroup$
      You did the wrong "plugging in". $3f(3x)-2f(2x)=3cdotfrac{(3x)^2-1}{(3x^2)^2+1}-2cdotfrac{(2x)^2-1}{(2x^2)^2+1}={color{red}{frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}}}$ is your answer.
      $endgroup$
      – user587192
      Dec 10 '18 at 13:25




















    • $begingroup$
      So after plugging in I get $frac{27x^2-3}{27x^2+3}-frac{8x^2-2}{8x^2+2}$ Then how do i arrive at the answer?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:53












    • $begingroup$
      You did the wrong "plugging in". $3f(3x)-2f(2x)=3cdotfrac{(3x)^2-1}{(3x^2)^2+1}-2cdotfrac{(2x)^2-1}{(2x^2)^2+1}={color{red}{frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}}}$ is your answer.
      $endgroup$
      – user587192
      Dec 10 '18 at 13:25


















    $begingroup$
    So after plugging in I get $frac{27x^2-3}{27x^2+3}-frac{8x^2-2}{8x^2+2}$ Then how do i arrive at the answer?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:53






    $begingroup$
    So after plugging in I get $frac{27x^2-3}{27x^2+3}-frac{8x^2-2}{8x^2+2}$ Then how do i arrive at the answer?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:53














    $begingroup$
    You did the wrong "plugging in". $3f(3x)-2f(2x)=3cdotfrac{(3x)^2-1}{(3x^2)^2+1}-2cdotfrac{(2x)^2-1}{(2x^2)^2+1}={color{red}{frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}}}$ is your answer.
    $endgroup$
    – user587192
    Dec 10 '18 at 13:25






    $begingroup$
    You did the wrong "plugging in". $3f(3x)-2f(2x)=3cdotfrac{(3x)^2-1}{(3x^2)^2+1}-2cdotfrac{(2x)^2-1}{(2x^2)^2+1}={color{red}{frac{3(9x^2-1)}{9x^2+1}-frac{2(4x^2-1)}{4x^2+1}}}$ is your answer.
    $endgroup$
    – user587192
    Dec 10 '18 at 13:25













    1












    $begingroup$

    Hint



    $$frac{d}{dx}int_{a(x)}^{b(x)}f(u) , du=f(b(x)) frac{d}{dx}(b(x))-f(a(x)) frac{d}{dx}(a(x))$$
    In your case $a(x)=2x, b(x)=3x$ and $f(u)=frac{u^2-1}{u^2+1}$. You don't need the derivatives of $f$ that you have computed.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      so I sub in 3x for u to get $3x^2$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 3:44










    • $begingroup$
      @EricBrown Actually $f(3x)=frac{(3x)^2-1}{(3x)^2+1}$.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 3:48












    • $begingroup$
      so after plugging in I get $frac{9x^2-1}{9x^2+1}(3)-frac{4x^2-1}{4x^2+1}(2)$. Then I multiply the two fractions to get $$frac{27x^2-3}{27x^3+3}-frac{8x^2-2}{8x^2+2}$$
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:00












    • $begingroup$
      @EricBrown Yes that's correct.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 4:02










    • $begingroup$
      Do I still need to integrate the two fractions?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:35
















    1












    $begingroup$

    Hint



    $$frac{d}{dx}int_{a(x)}^{b(x)}f(u) , du=f(b(x)) frac{d}{dx}(b(x))-f(a(x)) frac{d}{dx}(a(x))$$
    In your case $a(x)=2x, b(x)=3x$ and $f(u)=frac{u^2-1}{u^2+1}$. You don't need the derivatives of $f$ that you have computed.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      so I sub in 3x for u to get $3x^2$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 3:44










    • $begingroup$
      @EricBrown Actually $f(3x)=frac{(3x)^2-1}{(3x)^2+1}$.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 3:48












    • $begingroup$
      so after plugging in I get $frac{9x^2-1}{9x^2+1}(3)-frac{4x^2-1}{4x^2+1}(2)$. Then I multiply the two fractions to get $$frac{27x^2-3}{27x^3+3}-frac{8x^2-2}{8x^2+2}$$
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:00












    • $begingroup$
      @EricBrown Yes that's correct.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 4:02










    • $begingroup$
      Do I still need to integrate the two fractions?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:35














    1












    1








    1





    $begingroup$

    Hint



    $$frac{d}{dx}int_{a(x)}^{b(x)}f(u) , du=f(b(x)) frac{d}{dx}(b(x))-f(a(x)) frac{d}{dx}(a(x))$$
    In your case $a(x)=2x, b(x)=3x$ and $f(u)=frac{u^2-1}{u^2+1}$. You don't need the derivatives of $f$ that you have computed.






    share|cite|improve this answer









    $endgroup$



    Hint



    $$frac{d}{dx}int_{a(x)}^{b(x)}f(u) , du=f(b(x)) frac{d}{dx}(b(x))-f(a(x)) frac{d}{dx}(a(x))$$
    In your case $a(x)=2x, b(x)=3x$ and $f(u)=frac{u^2-1}{u^2+1}$. You don't need the derivatives of $f$ that you have computed.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 3:19









    Anurag AAnurag A

    26k12249




    26k12249












    • $begingroup$
      so I sub in 3x for u to get $3x^2$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 3:44










    • $begingroup$
      @EricBrown Actually $f(3x)=frac{(3x)^2-1}{(3x)^2+1}$.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 3:48












    • $begingroup$
      so after plugging in I get $frac{9x^2-1}{9x^2+1}(3)-frac{4x^2-1}{4x^2+1}(2)$. Then I multiply the two fractions to get $$frac{27x^2-3}{27x^3+3}-frac{8x^2-2}{8x^2+2}$$
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:00












    • $begingroup$
      @EricBrown Yes that's correct.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 4:02










    • $begingroup$
      Do I still need to integrate the two fractions?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:35


















    • $begingroup$
      so I sub in 3x for u to get $3x^2$?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 3:44










    • $begingroup$
      @EricBrown Actually $f(3x)=frac{(3x)^2-1}{(3x)^2+1}$.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 3:48












    • $begingroup$
      so after plugging in I get $frac{9x^2-1}{9x^2+1}(3)-frac{4x^2-1}{4x^2+1}(2)$. Then I multiply the two fractions to get $$frac{27x^2-3}{27x^3+3}-frac{8x^2-2}{8x^2+2}$$
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:00












    • $begingroup$
      @EricBrown Yes that's correct.
      $endgroup$
      – Anurag A
      Dec 10 '18 at 4:02










    • $begingroup$
      Do I still need to integrate the two fractions?
      $endgroup$
      – Eric Brown
      Dec 10 '18 at 4:35
















    $begingroup$
    so I sub in 3x for u to get $3x^2$?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 3:44




    $begingroup$
    so I sub in 3x for u to get $3x^2$?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 3:44












    $begingroup$
    @EricBrown Actually $f(3x)=frac{(3x)^2-1}{(3x)^2+1}$.
    $endgroup$
    – Anurag A
    Dec 10 '18 at 3:48






    $begingroup$
    @EricBrown Actually $f(3x)=frac{(3x)^2-1}{(3x)^2+1}$.
    $endgroup$
    – Anurag A
    Dec 10 '18 at 3:48














    $begingroup$
    so after plugging in I get $frac{9x^2-1}{9x^2+1}(3)-frac{4x^2-1}{4x^2+1}(2)$. Then I multiply the two fractions to get $$frac{27x^2-3}{27x^3+3}-frac{8x^2-2}{8x^2+2}$$
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:00






    $begingroup$
    so after plugging in I get $frac{9x^2-1}{9x^2+1}(3)-frac{4x^2-1}{4x^2+1}(2)$. Then I multiply the two fractions to get $$frac{27x^2-3}{27x^3+3}-frac{8x^2-2}{8x^2+2}$$
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:00














    $begingroup$
    @EricBrown Yes that's correct.
    $endgroup$
    – Anurag A
    Dec 10 '18 at 4:02




    $begingroup$
    @EricBrown Yes that's correct.
    $endgroup$
    – Anurag A
    Dec 10 '18 at 4:02












    $begingroup$
    Do I still need to integrate the two fractions?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:35




    $begingroup$
    Do I still need to integrate the two fractions?
    $endgroup$
    – Eric Brown
    Dec 10 '18 at 4:35











    1












    $begingroup$

    $$I(u)=intfrac{u^2-1}{u^2+1}mathrm du$$
    $$I(u)=intfrac{u^2+1}{u^2+1}mathrm du-2intfrac{mathrm du}{u^2+1}$$
    $$I(u)=u-2intfrac{mathrm du}{u^2+1}$$
    Using the substitution $tan t=u$, we arrive at
    $$I(u)=u-2arctan u$$
    And from $g(x)=I(3x)-I(2x)$,
    $$g(x)=x+arctan2x-arctan3x$$
    Hence
    $$g'(x)=1+frac2{1+4x^2}-frac3{1+9x^2}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$I(u)=intfrac{u^2-1}{u^2+1}mathrm du$$
      $$I(u)=intfrac{u^2+1}{u^2+1}mathrm du-2intfrac{mathrm du}{u^2+1}$$
      $$I(u)=u-2intfrac{mathrm du}{u^2+1}$$
      Using the substitution $tan t=u$, we arrive at
      $$I(u)=u-2arctan u$$
      And from $g(x)=I(3x)-I(2x)$,
      $$g(x)=x+arctan2x-arctan3x$$
      Hence
      $$g'(x)=1+frac2{1+4x^2}-frac3{1+9x^2}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$I(u)=intfrac{u^2-1}{u^2+1}mathrm du$$
        $$I(u)=intfrac{u^2+1}{u^2+1}mathrm du-2intfrac{mathrm du}{u^2+1}$$
        $$I(u)=u-2intfrac{mathrm du}{u^2+1}$$
        Using the substitution $tan t=u$, we arrive at
        $$I(u)=u-2arctan u$$
        And from $g(x)=I(3x)-I(2x)$,
        $$g(x)=x+arctan2x-arctan3x$$
        Hence
        $$g'(x)=1+frac2{1+4x^2}-frac3{1+9x^2}$$






        share|cite|improve this answer









        $endgroup$



        $$I(u)=intfrac{u^2-1}{u^2+1}mathrm du$$
        $$I(u)=intfrac{u^2+1}{u^2+1}mathrm du-2intfrac{mathrm du}{u^2+1}$$
        $$I(u)=u-2intfrac{mathrm du}{u^2+1}$$
        Using the substitution $tan t=u$, we arrive at
        $$I(u)=u-2arctan u$$
        And from $g(x)=I(3x)-I(2x)$,
        $$g(x)=x+arctan2x-arctan3x$$
        Hence
        $$g'(x)=1+frac2{1+4x^2}-frac3{1+9x^2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 3:24









        clathratusclathratus

        3,962334




        3,962334






























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