A slightly problematic integral $int{1/(x^4+1)^{1/4}} , mathrm{d}x$












9












$begingroup$


Question. To find the integral of- $$int {frac{1}{(x^4+1)^frac{1}{4}} , mathrm{d}x}$$



I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Wolfram gives me a difficult result, but it is in the form of $arctan$ and $ln$
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:20










  • $begingroup$
    @b00nheT, I am not familiar, regrettably, with hyperbolic functions. It isn't in our syllabus, either. We are to find it without those functions.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:23


















9












$begingroup$


Question. To find the integral of- $$int {frac{1}{(x^4+1)^frac{1}{4}} , mathrm{d}x}$$



I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Wolfram gives me a difficult result, but it is in the form of $arctan$ and $ln$
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:20










  • $begingroup$
    @b00nheT, I am not familiar, regrettably, with hyperbolic functions. It isn't in our syllabus, either. We are to find it without those functions.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:23
















9












9








9


1



$begingroup$


Question. To find the integral of- $$int {frac{1}{(x^4+1)^frac{1}{4}} , mathrm{d}x}$$



I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$




Question. To find the integral of- $$int {frac{1}{(x^4+1)^frac{1}{4}} , mathrm{d}x}$$



I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?






calculus integration indefinite-integrals




calculus integration indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 10:03







Nikhil Itty

















asked Jun 30 '16 at 9:15









Nikhil IttyNikhil Itty

1758




1758












  • $begingroup$
    Wolfram gives me a difficult result, but it is in the form of $arctan$ and $ln$
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:20










  • $begingroup$
    @b00nheT, I am not familiar, regrettably, with hyperbolic functions. It isn't in our syllabus, either. We are to find it without those functions.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:23




















  • $begingroup$
    Wolfram gives me a difficult result, but it is in the form of $arctan$ and $ln$
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:20










  • $begingroup$
    @b00nheT, I am not familiar, regrettably, with hyperbolic functions. It isn't in our syllabus, either. We are to find it without those functions.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:23


















$begingroup$
Wolfram gives me a difficult result, but it is in the form of $arctan$ and $ln$
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:20




$begingroup$
Wolfram gives me a difficult result, but it is in the form of $arctan$ and $ln$
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:20












$begingroup$
@b00nheT, I am not familiar, regrettably, with hyperbolic functions. It isn't in our syllabus, either. We are to find it without those functions.
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:23






$begingroup$
@b00nheT, I am not familiar, regrettably, with hyperbolic functions. It isn't in our syllabus, either. We are to find it without those functions.
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:23












2 Answers
2






active

oldest

votes


















17












$begingroup$

Hint:



This can be written as :
$$int frac{x^4dx}{x^5left(1+frac{1}{x^4}right)^{1/4}}$$
Now substitute $1+frac{1}{x^4}=t^4$
$$implies t^3dt=-frac{1}{x^5}dx$$
and
$$x^4=frac{1}{t^4-1}$$ to get
$$int frac{t^2dt}{1-t^4}$$
Now use partial fractions.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Just when I figured it out. Exactly one moment before. Thanks, mate.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:32



















9












$begingroup$

Let $$I = intfrac{1}{(x^4+1)^{frac{1}{4}}}dx$$



Put $x^2=tan theta,$ Then $2xdx = sec^2 theta dtheta$



So $$I = intfrac{sec^2 theta}{sqrt{sec theta}}cdot frac{1}{2sqrt{tan theta}}dtheta = frac{1}{2}intfrac{1}{cos theta sqrt{sin theta}}dtheta = frac{1}{2}intfrac{cos theta}{(1-sin^2 theta)sqrt{sin theta}}dtheta$$



Now Put $sin theta = t^2;,$ Then $cos theta dtheta = 2tdt$



So $$I = intfrac{1}{1-t^4}dt = -intfrac{1}{(t^2-1)(t^2+1)}dt = -frac{1}{2}intleft[frac{1}{1-t^2}+frac{1}{1+t^2}right]dt$$



So $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $frac{1}{2}$, otherwise our answers match.
    $endgroup$
    – Nikunj
    Jun 30 '16 at 11:47










  • $begingroup$
    Thanks Nikung, But your,s solution is much better then mine.
    $endgroup$
    – juantheron
    Jun 30 '16 at 12:16











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









17












$begingroup$

Hint:



This can be written as :
$$int frac{x^4dx}{x^5left(1+frac{1}{x^4}right)^{1/4}}$$
Now substitute $1+frac{1}{x^4}=t^4$
$$implies t^3dt=-frac{1}{x^5}dx$$
and
$$x^4=frac{1}{t^4-1}$$ to get
$$int frac{t^2dt}{1-t^4}$$
Now use partial fractions.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Just when I figured it out. Exactly one moment before. Thanks, mate.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:32
















17












$begingroup$

Hint:



This can be written as :
$$int frac{x^4dx}{x^5left(1+frac{1}{x^4}right)^{1/4}}$$
Now substitute $1+frac{1}{x^4}=t^4$
$$implies t^3dt=-frac{1}{x^5}dx$$
and
$$x^4=frac{1}{t^4-1}$$ to get
$$int frac{t^2dt}{1-t^4}$$
Now use partial fractions.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Just when I figured it out. Exactly one moment before. Thanks, mate.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:32














17












17








17





$begingroup$

Hint:



This can be written as :
$$int frac{x^4dx}{x^5left(1+frac{1}{x^4}right)^{1/4}}$$
Now substitute $1+frac{1}{x^4}=t^4$
$$implies t^3dt=-frac{1}{x^5}dx$$
and
$$x^4=frac{1}{t^4-1}$$ to get
$$int frac{t^2dt}{1-t^4}$$
Now use partial fractions.






share|cite|improve this answer











$endgroup$



Hint:



This can be written as :
$$int frac{x^4dx}{x^5left(1+frac{1}{x^4}right)^{1/4}}$$
Now substitute $1+frac{1}{x^4}=t^4$
$$implies t^3dt=-frac{1}{x^5}dx$$
and
$$x^4=frac{1}{t^4-1}$$ to get
$$int frac{t^2dt}{1-t^4}$$
Now use partial fractions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 30 '16 at 9:40

























answered Jun 30 '16 at 9:28









NikunjNikunj

4,50211233




4,50211233








  • 2




    $begingroup$
    Just when I figured it out. Exactly one moment before. Thanks, mate.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:32














  • 2




    $begingroup$
    Just when I figured it out. Exactly one moment before. Thanks, mate.
    $endgroup$
    – Nikhil Itty
    Jun 30 '16 at 9:32








2




2




$begingroup$
Just when I figured it out. Exactly one moment before. Thanks, mate.
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:32




$begingroup$
Just when I figured it out. Exactly one moment before. Thanks, mate.
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:32











9












$begingroup$

Let $$I = intfrac{1}{(x^4+1)^{frac{1}{4}}}dx$$



Put $x^2=tan theta,$ Then $2xdx = sec^2 theta dtheta$



So $$I = intfrac{sec^2 theta}{sqrt{sec theta}}cdot frac{1}{2sqrt{tan theta}}dtheta = frac{1}{2}intfrac{1}{cos theta sqrt{sin theta}}dtheta = frac{1}{2}intfrac{cos theta}{(1-sin^2 theta)sqrt{sin theta}}dtheta$$



Now Put $sin theta = t^2;,$ Then $cos theta dtheta = 2tdt$



So $$I = intfrac{1}{1-t^4}dt = -intfrac{1}{(t^2-1)(t^2+1)}dt = -frac{1}{2}intleft[frac{1}{1-t^2}+frac{1}{1+t^2}right]dt$$



So $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $frac{1}{2}$, otherwise our answers match.
    $endgroup$
    – Nikunj
    Jun 30 '16 at 11:47










  • $begingroup$
    Thanks Nikung, But your,s solution is much better then mine.
    $endgroup$
    – juantheron
    Jun 30 '16 at 12:16
















9












$begingroup$

Let $$I = intfrac{1}{(x^4+1)^{frac{1}{4}}}dx$$



Put $x^2=tan theta,$ Then $2xdx = sec^2 theta dtheta$



So $$I = intfrac{sec^2 theta}{sqrt{sec theta}}cdot frac{1}{2sqrt{tan theta}}dtheta = frac{1}{2}intfrac{1}{cos theta sqrt{sin theta}}dtheta = frac{1}{2}intfrac{cos theta}{(1-sin^2 theta)sqrt{sin theta}}dtheta$$



Now Put $sin theta = t^2;,$ Then $cos theta dtheta = 2tdt$



So $$I = intfrac{1}{1-t^4}dt = -intfrac{1}{(t^2-1)(t^2+1)}dt = -frac{1}{2}intleft[frac{1}{1-t^2}+frac{1}{1+t^2}right]dt$$



So $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $frac{1}{2}$, otherwise our answers match.
    $endgroup$
    – Nikunj
    Jun 30 '16 at 11:47










  • $begingroup$
    Thanks Nikung, But your,s solution is much better then mine.
    $endgroup$
    – juantheron
    Jun 30 '16 at 12:16














9












9








9





$begingroup$

Let $$I = intfrac{1}{(x^4+1)^{frac{1}{4}}}dx$$



Put $x^2=tan theta,$ Then $2xdx = sec^2 theta dtheta$



So $$I = intfrac{sec^2 theta}{sqrt{sec theta}}cdot frac{1}{2sqrt{tan theta}}dtheta = frac{1}{2}intfrac{1}{cos theta sqrt{sin theta}}dtheta = frac{1}{2}intfrac{cos theta}{(1-sin^2 theta)sqrt{sin theta}}dtheta$$



Now Put $sin theta = t^2;,$ Then $cos theta dtheta = 2tdt$



So $$I = intfrac{1}{1-t^4}dt = -intfrac{1}{(t^2-1)(t^2+1)}dt = -frac{1}{2}intleft[frac{1}{1-t^2}+frac{1}{1+t^2}right]dt$$



So $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$






share|cite|improve this answer











$endgroup$



Let $$I = intfrac{1}{(x^4+1)^{frac{1}{4}}}dx$$



Put $x^2=tan theta,$ Then $2xdx = sec^2 theta dtheta$



So $$I = intfrac{sec^2 theta}{sqrt{sec theta}}cdot frac{1}{2sqrt{tan theta}}dtheta = frac{1}{2}intfrac{1}{cos theta sqrt{sin theta}}dtheta = frac{1}{2}intfrac{cos theta}{(1-sin^2 theta)sqrt{sin theta}}dtheta$$



Now Put $sin theta = t^2;,$ Then $cos theta dtheta = 2tdt$



So $$I = intfrac{1}{1-t^4}dt = -intfrac{1}{(t^2-1)(t^2+1)}dt = -frac{1}{2}intleft[frac{1}{1-t^2}+frac{1}{1+t^2}right]dt$$



So $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 30 '16 at 12:16

























answered Jun 30 '16 at 10:28









juantheronjuantheron

34.3k1147142




34.3k1147142








  • 3




    $begingroup$
    Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $frac{1}{2}$, otherwise our answers match.
    $endgroup$
    – Nikunj
    Jun 30 '16 at 11:47










  • $begingroup$
    Thanks Nikung, But your,s solution is much better then mine.
    $endgroup$
    – juantheron
    Jun 30 '16 at 12:16














  • 3




    $begingroup$
    Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $frac{1}{2}$, otherwise our answers match.
    $endgroup$
    – Nikunj
    Jun 30 '16 at 11:47










  • $begingroup$
    Thanks Nikung, But your,s solution is much better then mine.
    $endgroup$
    – juantheron
    Jun 30 '16 at 12:16








3




3




$begingroup$
Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $frac{1}{2}$, otherwise our answers match.
$endgroup$
– Nikunj
Jun 30 '16 at 11:47




$begingroup$
Nice!(+1) but the $tan^{-1}t$ is missing a coefficient of $frac{1}{2}$, otherwise our answers match.
$endgroup$
– Nikunj
Jun 30 '16 at 11:47












$begingroup$
Thanks Nikung, But your,s solution is much better then mine.
$endgroup$
– juantheron
Jun 30 '16 at 12:16




$begingroup$
Thanks Nikung, But your,s solution is much better then mine.
$endgroup$
– juantheron
Jun 30 '16 at 12:16


















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