A slightly problematic integral $int{1/(x^4+1)^{1/4}} , mathrm{d}x$
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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
asked Jun 30 '16 at 9:15
Nikhil IttyNikhil Itty
1758
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add a comment |
$begingroup$
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
asked Jun 30 '16 at 9:15
Nikhil IttyNikhil Itty
1758
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jun 30 '16 at 9:15
Nikhil IttyNikhil Itty
1758
$endgroup$
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
asked Jun 30 '16 at 9:15
Nikhil IttyNikhil Itty
1758
asked Jun 30 '16 at 9:15
Nikhil IttyNikhil Itty
1758
edited Dec 2 '18 at 10:03
Nikhil Itty
asked Jun 30 '16 at 9:15
Nikhil IttyNikhil Itty
1758
asked Jun 30 '16 at 9:15
Nikhil IttyNikhil Itty
1758
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
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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.
answered Jun 30 '16 at 9:28
NikunjNikunj
4,50211233
$endgroup$
2
$begingroup$
Just when I figured it out. Exactly one moment before. Thanks, mate.
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:32
add a comment |
$begingroup$
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 $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$
answered Jun 30 '16 at 10:28
juantheronjuantheron
34.3k1147142
$endgroup$
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
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Thanks Nikung, But your,s solution is much better then mine.
$endgroup$
– juantheron
Jun 30 '16 at 12:16
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
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.
answered Jun 30 '16 at 9:28
NikunjNikunj
4,50211233
$endgroup$
2
$begingroup$
Just when I figured it out. Exactly one moment before. Thanks, mate.
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:32
add a comment |
$begingroup$
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.
answered Jun 30 '16 at 9:28
NikunjNikunj
4,50211233
$endgroup$
2
$begingroup$
Just when I figured it out. Exactly one moment before. Thanks, mate.
$endgroup$
– Nikhil Itty
Jun 30 '16 at 9:32
add a comment |
$begingroup$
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.
answered Jun 30 '16 at 9:28
NikunjNikunj
4,50211233
$endgroup$
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.
answered Jun 30 '16 at 9:28
NikunjNikunj
4,50211233
edited Jun 30 '16 at 9:40
answered Jun 30 '16 at 9:28
NikunjNikunj
4,50211233
answered Jun 30 '16 at 9:28
NikunjNikunj
4,50211233
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
add a comment |
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
add a comment |
$begingroup$
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 $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$
answered Jun 30 '16 at 10:28
juantheronjuantheron
34.3k1147142
$endgroup$
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
add a comment |
$begingroup$
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 $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$
answered Jun 30 '16 at 10:28
juantheronjuantheron
34.3k1147142
$endgroup$
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
add a comment |
$begingroup$
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 $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$
answered Jun 30 '16 at 10:28
juantheronjuantheron
34.3k1147142
$endgroup$
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 $$I = frac{1}{2}ln left|frac{t-1}{t+1}right|-frac{1}{2}tan^{-1}(t)+mathcal{C}$$
answered Jun 30 '16 at 10:28
juantheronjuantheron
34.3k1147142
edited Jun 30 '16 at 12:16
answered Jun 30 '16 at 10:28
juantheronjuantheron
34.3k1147142
answered Jun 30 '16 at 10:28
juantheronjuantheron
34.3k1147142
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
add a comment |
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
add a comment |
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$begingroup$
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