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How to solve $int :frac3x^2+1x^3+2dx$?
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up vote
0
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favorite
I am trying to solve the following integral:
$$int :frac3x^2+1x^3+2dx$$
I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?
calculus integration indefinite-integrals
edited Jul 23 at 21:00
Abcd
2,3361624
asked Jul 23 at 20:56
Francis
543
 |Â
show 1 more comment
up vote
0
down vote
favorite
I am trying to solve the following integral:
$$int :frac3x^2+1x^3+2dx$$
I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?
calculus integration indefinite-integrals
edited Jul 23 at 21:00
Abcd
2,3361624
asked Jul 23 at 20:56
Francis
543
The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02
But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03
3
You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04
Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13
This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve the following integral:
$$int :frac3x^2+1x^3+2dx$$
I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?
calculus integration indefinite-integrals
edited Jul 23 at 21:00
Abcd
2,3361624
asked Jul 23 at 20:56
Francis
543
I am trying to solve the following integral:
$$int :frac3x^2+1x^3+2dx$$
I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?
calculus integration indefinite-integrals
edited Jul 23 at 21:00
Abcd
2,3361624
asked Jul 23 at 20:56
Francis
543
edited Jul 23 at 21:00
Abcd
2,3361624
edited Jul 23 at 21:00
Abcd
2,3361624
edited Jul 23 at 21:00
Abcd
2,3361624
2,3361624
asked Jul 23 at 20:56
Francis
543
asked Jul 23 at 20:56
Francis
543
asked Jul 23 at 20:56
Francis
543
543
The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02
But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03
3
You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04
Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13
This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19
 |Â
show 1 more comment
The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02
But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03
3
You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04
Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13
This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19
The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02
The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02
But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03
But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03
3
3
You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04
You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04
Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13
Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13
This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19
This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19
 |Â
show 1 more comment
4 Answers
4
active
oldest
votes
up vote
0
down vote
accepted
Hint: write your integral in the form
$$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
– mrtaurho
Jul 23 at 21:11
I think here in this case is no other way, let me say i know no other way.
– Dr. Sonnhard Graubner
Jul 23 at 21:14
Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
– Francis
Jul 23 at 21:15
1
You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
– Reese
Jul 23 at 21:19
add a comment |Â
up vote
1
down vote
Wolfy gives
a number of forms,
the first of which is
$ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
+ dfraclog(2^2/3 x + 2)3 2^2/3
+ dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
+C
$
answered Jul 23 at 21:46
marty cohen
69.1k446122
add a comment |Â
up vote
0
down vote
Substitute $u=x^3$ and you can reduce it down to:
$$frac13intfracu^-frac23u+2du+intfrac1u+2du$$
I believe it can be taken from here
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
add a comment |Â
up vote
0
down vote
$$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
$$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
$$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
$$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
Now
$$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
$$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint: write your integral in the form
$$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
– mrtaurho
Jul 23 at 21:11
I think here in this case is no other way, let me say i know no other way.
– Dr. Sonnhard Graubner
Jul 23 at 21:14
Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
– Francis
Jul 23 at 21:15
1
You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
– Reese
Jul 23 at 21:19
add a comment |Â
up vote
0
down vote
accepted
Hint: write your integral in the form
$$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
– mrtaurho
Jul 23 at 21:11
I think here in this case is no other way, let me say i know no other way.
– Dr. Sonnhard Graubner
Jul 23 at 21:14
Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
– Francis
Jul 23 at 21:15
1
You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
– Reese
Jul 23 at 21:19
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint: write your integral in the form
$$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
Hint: write your integral in the form
$$int frac3x^2+1(x+sqrt[3]2)(x^2-sqrt[3]2x+2^2/3)dx$$
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
answered Jul 23 at 21:05
Dr. Sonnhard Graubner
66.7k32659
66.7k32659
I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
– mrtaurho
Jul 23 at 21:11
I think here in this case is no other way, let me say i know no other way.
– Dr. Sonnhard Graubner
Jul 23 at 21:14
Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
– Francis
Jul 23 at 21:15
1
You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
– Reese
Jul 23 at 21:19
add a comment |Â
I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
– mrtaurho
Jul 23 at 21:11
I think here in this case is no other way, let me say i know no other way.
– Dr. Sonnhard Graubner
Jul 23 at 21:14
Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
– Francis
Jul 23 at 21:15
1
You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
– Reese
Jul 23 at 21:19
I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
– mrtaurho
Jul 23 at 21:11
I guess you could apply partial fractions to this "new" but it seems to me that this gets really messy. Is there no other way?
– mrtaurho
Jul 23 at 21:11
I think here in this case is no other way, let me say i know no other way.
– Dr. Sonnhard Graubner
Jul 23 at 21:14
I think here in this case is no other way, let me say i know no other way.
– Dr. Sonnhard Graubner
Jul 23 at 21:14
Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
– Francis
Jul 23 at 21:15
Yes. Technically you can factor anything and apply partial fractions in the real numbers, but I was trying to avoid this. I suppose there truly is no other way. Thanks for your help!
– Francis
Jul 23 at 21:15
1
1
You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
– Reese
Jul 23 at 21:19
You can clean things up a little by doing a substitution $u = frac1sqrt[3]2x$ first; doesn't help a lot, but it makes the numbers better.
– Reese
Jul 23 at 21:19
add a comment |Â
up vote
1
down vote
Wolfy gives
a number of forms,
the first of which is
$ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
+ dfraclog(2^2/3 x + 2)3 2^2/3
+ dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
+C
$
answered Jul 23 at 21:46
marty cohen
69.1k446122
add a comment |Â
up vote
1
down vote
Wolfy gives
a number of forms,
the first of which is
$ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
+ dfraclog(2^2/3 x + 2)3 2^2/3
+ dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
+C
$
answered Jul 23 at 21:46
marty cohen
69.1k446122
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Wolfy gives
a number of forms,
the first of which is
$ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
+ dfraclog(2^2/3 x + 2)3 2^2/3
+ dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
+C
$
answered Jul 23 at 21:46
marty cohen
69.1k446122
Wolfy gives
a number of forms,
the first of which is
$ log(x^3 + 2) - dfraclog(2^1/3 x^2 - 2^2/3 x + 2)6 2^2/3
+ dfraclog(2^2/3 x + 2)3 2^2/3
+ dfractan^-1(frac2^2/3 x - 1sqrt3)2^2/3 sqrt3
+C
$
answered Jul 23 at 21:46
marty cohen
69.1k446122
answered Jul 23 at 21:46
marty cohen
69.1k446122
answered Jul 23 at 21:46
marty cohen
69.1k446122
answered Jul 23 at 21:46
marty cohen
69.1k446122
69.1k446122
add a comment |Â
add a comment |Â
up vote
0
down vote
Substitute $u=x^3$ and you can reduce it down to:
$$frac13intfracu^-frac23u+2du+intfrac1u+2du$$
I believe it can be taken from here
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
add a comment |Â
up vote
0
down vote
Substitute $u=x^3$ and you can reduce it down to:
$$frac13intfracu^-frac23u+2du+intfrac1u+2du$$
I believe it can be taken from here
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Substitute $u=x^3$ and you can reduce it down to:
$$frac13intfracu^-frac23u+2du+intfrac1u+2du$$
I believe it can be taken from here
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
Substitute $u=x^3$ and you can reduce it down to:
$$frac13intfracu^-frac23u+2du+intfrac1u+2du$$
I believe it can be taken from here
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
answered Jul 23 at 21:19
Rhys Hughes
3,8681227
3,8681227
add a comment |Â
add a comment |Â
up vote
0
down vote
$$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
$$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
$$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
$$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
Now
$$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
$$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
$$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
$$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
$$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
$$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
Now
$$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
$$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
$$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
$$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
$$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
Now
$$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
$$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
$$I=int frac3x^2+1x^3+2dx=int frac3x^2x^3+2dx+int fracdxx^3+2$$
$$I=log(x^3+2)+int fracdxx^3+2$$ For the remaining integral, let
$$x=sqrt[3]2, timplies dx=sqrt[3]2,dt$$
$$J=int fracdxx^3+2=fracsqrt[3]22 int fracdtt^3+1$$
Now
$$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition
$$frac 1 t^3+1=frac1(a-1) (a-b) (t-a)+frac1(b-1) (b-a) (t-b)+frac1(a-1) (b-1) (t-1)$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
answered Jul 24 at 5:30
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
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The easy part is $frac3x^2x^3+2$ indeed $(log x^3+2)'=frac3x^2x^3+2$.
– gimusi
Jul 23 at 21:02
But I guess that does not help at all because the remaining integral would be $ int frac1x^3+2mathrmdx$ and I am not sure how to solve this.
– mrtaurho
Jul 23 at 21:03
3
You can factor any polynomial over $mathbbR$ into linear or 2nd degree polynomials. Here,you can factor $x^3+2=(x+sqrt[3] 2)(x^2-sqrt[3]2x+sqrt[3]4)$, if its any help.
– Sar
Jul 23 at 21:04
Separate the two and in the second integral take sub $1+2/x^3=t$
– Pi_die_die
Jul 23 at 21:13
This is it ! You can combine gimusi's trick and Dr. Sonnhard Graubner's answer. The linear and quadratic factors are mutually prime so you get $$frac1x^3+2=fracAx+sqrt[3]2+fracBx+Cx^2-sqrt[3]2x+2^2/3$$ the second can be written $$fraclambda Q'Q+fracmuQ$$ where $$Q(x)=x^2-sqrt[3]2x+2^2/3$$
– Duchamp Gérard H. E.
Jul 23 at 21:19