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How to calculate the indefinite integral of $frac1x^2/3(1+x^2/3)$?
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$$int fracdxx^2/3(1+x^2/3).$$
I substituted,
$$t=frac1x^1/3$$
$$fracdtdx = -frac13x^4/3$$
$$fracdtdx = -fract^43$$
Rewriting the question,
$$int fracdxx^2/3+x^4/3$$
$$-frac13 int fracdtt^4Bigl(frac1t^2+frac1t^4Bigr)$$
We have,
$$-frac13 int fracdtt^2 + 1$$
$$-frac13tan^-1t+C$$
$$-frac13tan^-1Biggl(frac1x^1/3Biggr)+C$$
But the answer given is $$3tan^-1x^1/3+C$$
Where am I wrong?
Any help would be appreciated.
integration trigonometry indefinite-integrals
edited Jul 26 at 13:47
AccidentalFourierTransform
1,075625
asked Jul 26 at 11:04
Piano Land
35812
add a comment |Â
up vote
4
down vote
favorite
$$int fracdxx^2/3(1+x^2/3).$$
I substituted,
$$t=frac1x^1/3$$
$$fracdtdx = -frac13x^4/3$$
$$fracdtdx = -fract^43$$
Rewriting the question,
$$int fracdxx^2/3+x^4/3$$
$$-frac13 int fracdtt^4Bigl(frac1t^2+frac1t^4Bigr)$$
We have,
$$-frac13 int fracdtt^2 + 1$$
$$-frac13tan^-1t+C$$
$$-frac13tan^-1Biggl(frac1x^1/3Biggr)+C$$
But the answer given is $$3tan^-1x^1/3+C$$
Where am I wrong?
Any help would be appreciated.
integration trigonometry indefinite-integrals
edited Jul 26 at 13:47
AccidentalFourierTransform
1,075625
asked Jul 26 at 11:04
Piano Land
35812
2
You forgot the number 3 to multiply the dt
– dmtri
Jul 26 at 11:15
In the integral, I mean
– dmtri
Jul 26 at 11:16
And not 1/3 as you have
– dmtri
Jul 26 at 11:18
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$$int fracdxx^2/3(1+x^2/3).$$
I substituted,
$$t=frac1x^1/3$$
$$fracdtdx = -frac13x^4/3$$
$$fracdtdx = -fract^43$$
Rewriting the question,
$$int fracdxx^2/3+x^4/3$$
$$-frac13 int fracdtt^4Bigl(frac1t^2+frac1t^4Bigr)$$
We have,
$$-frac13 int fracdtt^2 + 1$$
$$-frac13tan^-1t+C$$
$$-frac13tan^-1Biggl(frac1x^1/3Biggr)+C$$
But the answer given is $$3tan^-1x^1/3+C$$
Where am I wrong?
Any help would be appreciated.
integration trigonometry indefinite-integrals
edited Jul 26 at 13:47
AccidentalFourierTransform
1,075625
asked Jul 26 at 11:04
Piano Land
35812
$$int fracdxx^2/3(1+x^2/3).$$
I substituted,
$$t=frac1x^1/3$$
$$fracdtdx = -frac13x^4/3$$
$$fracdtdx = -fract^43$$
Rewriting the question,
$$int fracdxx^2/3+x^4/3$$
$$-frac13 int fracdtt^4Bigl(frac1t^2+frac1t^4Bigr)$$
We have,
$$-frac13 int fracdtt^2 + 1$$
$$-frac13tan^-1t+C$$
$$-frac13tan^-1Biggl(frac1x^1/3Biggr)+C$$
But the answer given is $$3tan^-1x^1/3+C$$
Where am I wrong?
Any help would be appreciated.
integration trigonometry indefinite-integrals
edited Jul 26 at 13:47
AccidentalFourierTransform
1,075625
asked Jul 26 at 11:04
Piano Land
35812
edited Jul 26 at 13:47
AccidentalFourierTransform
1,075625
edited Jul 26 at 13:47
AccidentalFourierTransform
1,075625
edited Jul 26 at 13:47
AccidentalFourierTransform
1,075625
1,075625
asked Jul 26 at 11:04
Piano Land
35812
asked Jul 26 at 11:04
Piano Land
35812
asked Jul 26 at 11:04
Piano Land
35812
35812
2
You forgot the number 3 to multiply the dt
– dmtri
Jul 26 at 11:15
In the integral, I mean
– dmtri
Jul 26 at 11:16
And not 1/3 as you have
– dmtri
Jul 26 at 11:18
add a comment |Â
2
You forgot the number 3 to multiply the dt
– dmtri
Jul 26 at 11:15
In the integral, I mean
– dmtri
Jul 26 at 11:16
And not 1/3 as you have
– dmtri
Jul 26 at 11:18
2
2
You forgot the number 3 to multiply the dt
– dmtri
Jul 26 at 11:15
You forgot the number 3 to multiply the dt
– dmtri
Jul 26 at 11:15
In the integral, I mean
– dmtri
Jul 26 at 11:16
In the integral, I mean
– dmtri
Jul 26 at 11:16
And not 1/3 as you have
– dmtri
Jul 26 at 11:18
And not 1/3 as you have
– dmtri
Jul 26 at 11:18
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
Two observations resolve this. The first is that you didn't invert the derivative properly, so your $1/3$ coefficient should be $3$ because $dx=-3t^-4dt$. Secondly, $arctan 1/y=pi/2-arctan y$ implies there's more than one way to write the answer, with your method getting something valid. I think you were expected to substitute $t=x^1/3$ instead; your choice does work, but it's conceptually more complex.
answered Jul 26 at 11:19
J.G.
13.1k11423
add a comment |Â
up vote
6
down vote
$t=sqrt[3]x,qquad dt=frac1x^2/3dx$
$$int frac1x^frac23+x^frac43dx=3int frac1t^2+1dt$$
answered Jul 26 at 11:19
Amarildo
1,724616
add a comment |Â
up vote
0
down vote
If $$t=x^-2/3$$ then $$dt=frac-23x^-5/3dx$$
edited Jul 26 at 11:57
peterh
2,14731631
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Two observations resolve this. The first is that you didn't invert the derivative properly, so your $1/3$ coefficient should be $3$ because $dx=-3t^-4dt$. Secondly, $arctan 1/y=pi/2-arctan y$ implies there's more than one way to write the answer, with your method getting something valid. I think you were expected to substitute $t=x^1/3$ instead; your choice does work, but it's conceptually more complex.
answered Jul 26 at 11:19
J.G.
13.1k11423
add a comment |Â
up vote
7
down vote
accepted
Two observations resolve this. The first is that you didn't invert the derivative properly, so your $1/3$ coefficient should be $3$ because $dx=-3t^-4dt$. Secondly, $arctan 1/y=pi/2-arctan y$ implies there's more than one way to write the answer, with your method getting something valid. I think you were expected to substitute $t=x^1/3$ instead; your choice does work, but it's conceptually more complex.
answered Jul 26 at 11:19
J.G.
13.1k11423
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Two observations resolve this. The first is that you didn't invert the derivative properly, so your $1/3$ coefficient should be $3$ because $dx=-3t^-4dt$. Secondly, $arctan 1/y=pi/2-arctan y$ implies there's more than one way to write the answer, with your method getting something valid. I think you were expected to substitute $t=x^1/3$ instead; your choice does work, but it's conceptually more complex.
answered Jul 26 at 11:19
J.G.
13.1k11423
Two observations resolve this. The first is that you didn't invert the derivative properly, so your $1/3$ coefficient should be $3$ because $dx=-3t^-4dt$. Secondly, $arctan 1/y=pi/2-arctan y$ implies there's more than one way to write the answer, with your method getting something valid. I think you were expected to substitute $t=x^1/3$ instead; your choice does work, but it's conceptually more complex.
answered Jul 26 at 11:19
J.G.
13.1k11423
answered Jul 26 at 11:19
J.G.
13.1k11423
answered Jul 26 at 11:19
J.G.
13.1k11423
answered Jul 26 at 11:19
J.G.
13.1k11423
13.1k11423
add a comment |Â
add a comment |Â
up vote
6
down vote
$t=sqrt[3]x,qquad dt=frac1x^2/3dx$
$$int frac1x^frac23+x^frac43dx=3int frac1t^2+1dt$$
answered Jul 26 at 11:19
Amarildo
1,724616
add a comment |Â
up vote
6
down vote
$t=sqrt[3]x,qquad dt=frac1x^2/3dx$
$$int frac1x^frac23+x^frac43dx=3int frac1t^2+1dt$$
answered Jul 26 at 11:19
Amarildo
1,724616
add a comment |Â
up vote
6
down vote
up vote
6
down vote
$t=sqrt[3]x,qquad dt=frac1x^2/3dx$
$$int frac1x^frac23+x^frac43dx=3int frac1t^2+1dt$$
answered Jul 26 at 11:19
Amarildo
1,724616
$t=sqrt[3]x,qquad dt=frac1x^2/3dx$
$$int frac1x^frac23+x^frac43dx=3int frac1t^2+1dt$$
answered Jul 26 at 11:19
Amarildo
1,724616
edited Jul 26 at 11:22
answered Jul 26 at 11:19
Amarildo
1,724616
answered Jul 26 at 11:19
Amarildo
1,724616
answered Jul 26 at 11:19
Amarildo
1,724616
1,724616
add a comment |Â
add a comment |Â
up vote
0
down vote
If $$t=x^-2/3$$ then $$dt=frac-23x^-5/3dx$$
edited Jul 26 at 11:57
peterh
2,14731631
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
up vote
0
down vote
If $$t=x^-2/3$$ then $$dt=frac-23x^-5/3dx$$
edited Jul 26 at 11:57
peterh
2,14731631
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $$t=x^-2/3$$ then $$dt=frac-23x^-5/3dx$$
edited Jul 26 at 11:57
peterh
2,14731631
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
If $$t=x^-2/3$$ then $$dt=frac-23x^-5/3dx$$
edited Jul 26 at 11:57
peterh
2,14731631
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
edited Jul 26 at 11:57
peterh
2,14731631
edited Jul 26 at 11:57
peterh
2,14731631
edited Jul 26 at 11:57
peterh
2,14731631
2,14731631
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
answered Jul 26 at 11:18
Dr. Sonnhard Graubner
66.7k32659
66.7k32659
add a comment |Â
add a comment |Â
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2
You forgot the number 3 to multiply the dt
– dmtri
Jul 26 at 11:15
In the integral, I mean
– dmtri
Jul 26 at 11:16
And not 1/3 as you have
– dmtri
Jul 26 at 11:18