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Evaluate the intgral $intfracdxx^2(1-x^2)$ (solution verification)
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up vote
3
down vote
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I have to solve the following integral
$$intfracdxx^2(1-x^2)$$
What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit
Is this correct?
Thanks in advance!
calculus integration proof-verification
edited Aug 2 at 14:57
John Ma
37.5k93669
asked Aug 2 at 11:31
Nikola
690617
add a comment |Â
up vote
3
down vote
favorite
I have to solve the following integral
$$intfracdxx^2(1-x^2)$$
What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit
Is this correct?
Thanks in advance!
calculus integration proof-verification
edited Aug 2 at 14:57
John Ma
37.5k93669
asked Aug 2 at 11:31
Nikola
690617
1
I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35
1
As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57
I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have to solve the following integral
$$intfracdxx^2(1-x^2)$$
What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit
Is this correct?
Thanks in advance!
calculus integration proof-verification
edited Aug 2 at 14:57
John Ma
37.5k93669
asked Aug 2 at 11:31
Nikola
690617
I have to solve the following integral
$$intfracdxx^2(1-x^2)$$
What I've got:
beginsplit
intfracdxx^2(1-x^2) &=intfrac(1-x^2+x^2)dxx^2(1-x^2)\
&=intfracdxx^2+intfracdx1-x^2\
&=intfracdxx^2+intfracdx(1+(xi)^2)\
&=-x^-1+arctanxi+C
endsplit
Is this correct?
Thanks in advance!
calculus integration proof-verification
edited Aug 2 at 14:57
John Ma
37.5k93669
asked Aug 2 at 11:31
Nikola
690617
edited Aug 2 at 14:57
John Ma
37.5k93669
edited Aug 2 at 14:57
John Ma
37.5k93669
edited Aug 2 at 14:57
John Ma
37.5k93669
37.5k93669
asked Aug 2 at 11:31
Nikola
690617
asked Aug 2 at 11:31
Nikola
690617
asked Aug 2 at 11:31
Nikola
690617
690617
1
I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35
1
As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57
I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58
add a comment |Â
1
I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35
1
As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57
I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58
1
1
I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35
I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35
1
1
As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57
As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57
I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58
I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
3
down vote
accepted
Don't forget $dfrac1i$ before $arctan$
$$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
Also better to write
$$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$
answered Aug 2 at 11:35
user 108128
18.8k41544
What problems there might be using the imaginary part? Is it a matter of preference?
– Nikola
Aug 2 at 11:51
Well, you say "its better to write". Why is that?
– Nikola
Aug 2 at 11:57
1
In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
– user 108128
Aug 2 at 12:02
add a comment |Â
up vote
3
down vote
Another method using the self-similar substitution
$$x=frac1-t1+t$$
transforms the integral to
$$-intfrac2left( t-1 right)^2+frac12tdt$$
answered Aug 2 at 22:23
Vincent Law
1,192110
add a comment |Â
up vote
2
down vote
Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
add a comment |Â
up vote
2
down vote
$x = tanh(u)$ then $dx = operatornamesech^2(u) du$
$$
beginalign
int fracdxx^2(1-x^2)
&= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
&= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
&= int fracdutanh^2(u)\
&= int fraccosh^2(u)sinh^2(u) du\
&= int (1 + operatornamecsch^2(u))du\
&= u - coth(u)\
&= operatornamearctanh(x) - frac1x + C
endalign
$$
edited Aug 2 at 12:43
user 108128
18.8k41544
answered Aug 2 at 12:27
Cato
1,09038
I've edited yours, if you don't want, so do roll back.
– user 108128
Aug 2 at 12:44
I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
– Cato
Aug 2 at 13:00
add a comment |Â
up vote
1
down vote
Note that, if $f(x)=arctan(xi)$, then
$$
f'(x)=ifrac11+(xi)^2=fraci1-x^2
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-frac1x-iarctan(xi)
$$
I'd prefer to avoid complex functions. With partial fractions, we set
$$
frac1x^2(1-x)^2=
fracax+fracbx^2+fraccx-1+fracdx+1
$$
Removing the denominators leads to
$$
ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
$$
With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.
Therefore
$$
intfrac1x^2(1-x^2),dx=
intleft(
frac1x^2-frac12frac1x-1+frac12frac1x+1
right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
$$
answered Aug 2 at 13:19
egreg
164k1180187
+1. Good illustration. end of solution is $frac12$ behind $log$.
– user 108128
Aug 2 at 13:38
@user108128 Thanks for noting
– egreg
Aug 2 at 14:14
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Don't forget $dfrac1i$ before $arctan$
$$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
Also better to write
$$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$
answered Aug 2 at 11:35
user 108128
18.8k41544
What problems there might be using the imaginary part? Is it a matter of preference?
– Nikola
Aug 2 at 11:51
Well, you say "its better to write". Why is that?
– Nikola
Aug 2 at 11:57
1
In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
– user 108128
Aug 2 at 12:02
add a comment |Â
up vote
3
down vote
accepted
Don't forget $dfrac1i$ before $arctan$
$$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
Also better to write
$$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$
answered Aug 2 at 11:35
user 108128
18.8k41544
What problems there might be using the imaginary part? Is it a matter of preference?
– Nikola
Aug 2 at 11:51
Well, you say "its better to write". Why is that?
– Nikola
Aug 2 at 11:57
1
In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
– user 108128
Aug 2 at 12:02
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Don't forget $dfrac1i$ before $arctan$
$$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
Also better to write
$$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$
answered Aug 2 at 11:35
user 108128
18.8k41544
Don't forget $dfrac1i$ before $arctan$
$$intfracdxx^2(1-x^2)=-x^-1+dfrac1iarctanxi+C$$
Also better to write
$$intdfrac11-x^2dx=dfrac12lndfrac1+x1-x+C$$
answered Aug 2 at 11:35
user 108128
18.8k41544
answered Aug 2 at 11:35
user 108128
18.8k41544
answered Aug 2 at 11:35
user 108128
18.8k41544
answered Aug 2 at 11:35
user 108128
18.8k41544
18.8k41544
What problems there might be using the imaginary part? Is it a matter of preference?
– Nikola
Aug 2 at 11:51
Well, you say "its better to write". Why is that?
– Nikola
Aug 2 at 11:57
1
In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
– user 108128
Aug 2 at 12:02
add a comment |Â
What problems there might be using the imaginary part? Is it a matter of preference?
– Nikola
Aug 2 at 11:51
Well, you say "its better to write". Why is that?
– Nikola
Aug 2 at 11:57
1
In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
– user 108128
Aug 2 at 12:02
What problems there might be using the imaginary part? Is it a matter of preference?
– Nikola
Aug 2 at 11:51
What problems there might be using the imaginary part? Is it a matter of preference?
– Nikola
Aug 2 at 11:51
Well, you say "its better to write". Why is that?
– Nikola
Aug 2 at 11:57
Well, you say "its better to write". Why is that?
– Nikola
Aug 2 at 11:57
1
1
In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
– user 108128
Aug 2 at 12:02
In real analysis we should avoid using imaginary part as possible, and it's better use functions with one variable $x$, and leave $i$ for complex analyysis functions.
– user 108128
Aug 2 at 12:02
add a comment |Â
up vote
3
down vote
Another method using the self-similar substitution
$$x=frac1-t1+t$$
transforms the integral to
$$-intfrac2left( t-1 right)^2+frac12tdt$$
answered Aug 2 at 22:23
Vincent Law
1,192110
add a comment |Â
up vote
3
down vote
Another method using the self-similar substitution
$$x=frac1-t1+t$$
transforms the integral to
$$-intfrac2left( t-1 right)^2+frac12tdt$$
answered Aug 2 at 22:23
Vincent Law
1,192110
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Another method using the self-similar substitution
$$x=frac1-t1+t$$
transforms the integral to
$$-intfrac2left( t-1 right)^2+frac12tdt$$
answered Aug 2 at 22:23
Vincent Law
1,192110
Another method using the self-similar substitution
$$x=frac1-t1+t$$
transforms the integral to
$$-intfrac2left( t-1 right)^2+frac12tdt$$
answered Aug 2 at 22:23
Vincent Law
1,192110
edited Aug 2 at 22:38
answered Aug 2 at 22:23
Vincent Law
1,192110
answered Aug 2 at 22:23
Vincent Law
1,192110
answered Aug 2 at 22:23
Vincent Law
1,192110
1,192110
add a comment |Â
add a comment |Â
up vote
2
down vote
Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
add a comment |Â
up vote
2
down vote
Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $frac11-x^2$ should have been $-iarctan(xi)$. But I suggest that you do$$frac11-x^2=frac1/21-x+frac1/21+x$$instead.
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
answered Aug 2 at 11:35
José Carlos Santos
112k1696172
112k1696172
add a comment |Â
add a comment |Â
up vote
2
down vote
$x = tanh(u)$ then $dx = operatornamesech^2(u) du$
$$
beginalign
int fracdxx^2(1-x^2)
&= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
&= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
&= int fracdutanh^2(u)\
&= int fraccosh^2(u)sinh^2(u) du\
&= int (1 + operatornamecsch^2(u))du\
&= u - coth(u)\
&= operatornamearctanh(x) - frac1x + C
endalign
$$
edited Aug 2 at 12:43
user 108128
18.8k41544
answered Aug 2 at 12:27
Cato
1,09038
I've edited yours, if you don't want, so do roll back.
– user 108128
Aug 2 at 12:44
I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
– Cato
Aug 2 at 13:00
add a comment |Â
up vote
2
down vote
$x = tanh(u)$ then $dx = operatornamesech^2(u) du$
$$
beginalign
int fracdxx^2(1-x^2)
&= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
&= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
&= int fracdutanh^2(u)\
&= int fraccosh^2(u)sinh^2(u) du\
&= int (1 + operatornamecsch^2(u))du\
&= u - coth(u)\
&= operatornamearctanh(x) - frac1x + C
endalign
$$
edited Aug 2 at 12:43
user 108128
18.8k41544
answered Aug 2 at 12:27
Cato
1,09038
I've edited yours, if you don't want, so do roll back.
– user 108128
Aug 2 at 12:44
I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
– Cato
Aug 2 at 13:00
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$x = tanh(u)$ then $dx = operatornamesech^2(u) du$
$$
beginalign
int fracdxx^2(1-x^2)
&= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
&= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
&= int fracdutanh^2(u)\
&= int fraccosh^2(u)sinh^2(u) du\
&= int (1 + operatornamecsch^2(u))du\
&= u - coth(u)\
&= operatornamearctanh(x) - frac1x + C
endalign
$$
edited Aug 2 at 12:43
user 108128
18.8k41544
answered Aug 2 at 12:27
Cato
1,09038
$x = tanh(u)$ then $dx = operatornamesech^2(u) du$
$$
beginalign
int fracdxx^2(1-x^2)
&= int fracoperatornamesech^2(u) dutanh^2(u)(1-tanh^2(u))\
&= int fracoperatornamesech^2(u) dutanh^2(u)(operatornamesech^2(u))\
&= int fracdutanh^2(u)\
&= int fraccosh^2(u)sinh^2(u) du\
&= int (1 + operatornamecsch^2(u))du\
&= u - coth(u)\
&= operatornamearctanh(x) - frac1x + C
endalign
$$
edited Aug 2 at 12:43
user 108128
18.8k41544
answered Aug 2 at 12:27
Cato
1,09038
edited Aug 2 at 12:43
user 108128
18.8k41544
edited Aug 2 at 12:43
user 108128
18.8k41544
edited Aug 2 at 12:43
user 108128
18.8k41544
18.8k41544
answered Aug 2 at 12:27
Cato
1,09038
answered Aug 2 at 12:27
Cato
1,09038
answered Aug 2 at 12:27
Cato
1,09038
1,09038
I've edited yours, if you don't want, so do roll back.
– user 108128
Aug 2 at 12:44
I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
– Cato
Aug 2 at 13:00
add a comment |Â
I've edited yours, if you don't want, so do roll back.
– user 108128
Aug 2 at 12:44
I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
– Cato
Aug 2 at 13:00
I've edited yours, if you don't want, so do roll back.
– user 108128
Aug 2 at 12:44
I've edited yours, if you don't want, so do roll back.
– user 108128
Aug 2 at 12:44
I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
– Cato
Aug 2 at 13:00
I think the original questioners partial fraction method was better - apart from missing the factorization of $1 - x^2$ - it was probably easier. I once integrated $frac11 - x^2$ the same sort of way and handed it in to someone
– Cato
Aug 2 at 13:00
add a comment |Â
up vote
1
down vote
Note that, if $f(x)=arctan(xi)$, then
$$
f'(x)=ifrac11+(xi)^2=fraci1-x^2
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-frac1x-iarctan(xi)
$$
I'd prefer to avoid complex functions. With partial fractions, we set
$$
frac1x^2(1-x)^2=
fracax+fracbx^2+fraccx-1+fracdx+1
$$
Removing the denominators leads to
$$
ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
$$
With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.
Therefore
$$
intfrac1x^2(1-x^2),dx=
intleft(
frac1x^2-frac12frac1x-1+frac12frac1x+1
right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
$$
answered Aug 2 at 13:19
egreg
164k1180187
+1. Good illustration. end of solution is $frac12$ behind $log$.
– user 108128
Aug 2 at 13:38
@user108128 Thanks for noting
– egreg
Aug 2 at 14:14
add a comment |Â
up vote
1
down vote
Note that, if $f(x)=arctan(xi)$, then
$$
f'(x)=ifrac11+(xi)^2=fraci1-x^2
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-frac1x-iarctan(xi)
$$
I'd prefer to avoid complex functions. With partial fractions, we set
$$
frac1x^2(1-x)^2=
fracax+fracbx^2+fraccx-1+fracdx+1
$$
Removing the denominators leads to
$$
ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
$$
With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.
Therefore
$$
intfrac1x^2(1-x^2),dx=
intleft(
frac1x^2-frac12frac1x-1+frac12frac1x+1
right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
$$
answered Aug 2 at 13:19
egreg
164k1180187
+1. Good illustration. end of solution is $frac12$ behind $log$.
– user 108128
Aug 2 at 13:38
@user108128 Thanks for noting
– egreg
Aug 2 at 14:14
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that, if $f(x)=arctan(xi)$, then
$$
f'(x)=ifrac11+(xi)^2=fraci1-x^2
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-frac1x-iarctan(xi)
$$
I'd prefer to avoid complex functions. With partial fractions, we set
$$
frac1x^2(1-x)^2=
fracax+fracbx^2+fraccx-1+fracdx+1
$$
Removing the denominators leads to
$$
ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
$$
With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.
Therefore
$$
intfrac1x^2(1-x^2),dx=
intleft(
frac1x^2-frac12frac1x-1+frac12frac1x+1
right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
$$
answered Aug 2 at 13:19
egreg
164k1180187
Note that, if $f(x)=arctan(xi)$, then
$$
f'(x)=ifrac11+(xi)^2=fraci1-x^2
$$
that's not the required derivative; you are missing $1/i=-i$, so you could write
$$
-frac1x-iarctan(xi)
$$
I'd prefer to avoid complex functions. With partial fractions, we set
$$
frac1x^2(1-x)^2=
fracax+fracbx^2+fraccx-1+fracdx+1
$$
Removing the denominators leads to
$$
ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1
$$
With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.
Therefore
$$
intfrac1x^2(1-x^2),dx=
intleft(
frac1x^2-frac12frac1x-1+frac12frac1x+1
right),dx=-frac1x+frac12logleft|fracx+1x-1right|+c
$$
answered Aug 2 at 13:19
egreg
164k1180187
edited Aug 2 at 14:14
answered Aug 2 at 13:19
egreg
164k1180187
answered Aug 2 at 13:19
egreg
164k1180187
answered Aug 2 at 13:19
egreg
164k1180187
164k1180187
+1. Good illustration. end of solution is $frac12$ behind $log$.
– user 108128
Aug 2 at 13:38
@user108128 Thanks for noting
– egreg
Aug 2 at 14:14
add a comment |Â
+1. Good illustration. end of solution is $frac12$ behind $log$.
– user 108128
Aug 2 at 13:38
@user108128 Thanks for noting
– egreg
Aug 2 at 14:14
+1. Good illustration. end of solution is $frac12$ behind $log$.
– user 108128
Aug 2 at 13:38
+1. Good illustration. end of solution is $frac12$ behind $log$.
– user 108128
Aug 2 at 13:38
@user108128 Thanks for noting
– egreg
Aug 2 at 14:14
@user108128 Thanks for noting
– egreg
Aug 2 at 14:14
add a comment |Â
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1
I would suggest partial fraction decomposition for the last integral instead of setting $x=xi$. You cannot just change a variable without changing the differential too but it should work aswell.
– mrtaurho
Aug 2 at 11:35
1
As user 108128 already noted, you missed a factor of $-i$ on the arctangent. In fact we can write $-ixarctan ix$ as a hyperbolic arctangent, $operatornameartanh x$.
– J.G.
Aug 2 at 11:57
I think it's good that you came up with your own method, but you probably went the wrong way on the $frac11 - x^2$ part, which can be done different ways google.com/…
– Cato
Aug 2 at 12:58