Mixing
Integration by partial fractions problem
$begingroup$
I'm having a lot of trouble with this integral. I don't know what to set A+B equal to. There's an x^4 in the numerator and I'm trying to figure out how to account for it. What am I doing wrong? Thanks.
Problem: $$intfrac{x^4}{4-x^2}space dx$$
What I have tried:
$$frac{x^4}{(2-x)(2+x)}=frac{a}{2-x}+frac{b}{2+x}=frac{a(2+x)+b(2-x)}{(2-x)(2+x)}implies x^4=(a+b)x+2(a+b)$$
integration partial-fractions
edited Jan 18 at 0:02
coreyman317
774420
asked Jan 17 at 23:06
J.W.J.W.
524
$endgroup$
add a comment |
$begingroup$
I'm having a lot of trouble with this integral. I don't know what to set A+B equal to. There's an x^4 in the numerator and I'm trying to figure out how to account for it. What am I doing wrong? Thanks.
Problem: $$intfrac{x^4}{4-x^2}space dx$$
What I have tried:
$$frac{x^4}{(2-x)(2+x)}=frac{a}{2-x}+frac{b}{2+x}=frac{a(2+x)+b(2-x)}{(2-x)(2+x)}implies x^4=(a+b)x+2(a+b)$$
integration partial-fractions
edited Jan 18 at 0:02
coreyman317
774420
asked Jan 17 at 23:06
J.W.J.W.
524
$endgroup$
1
$begingroup$
First of all, do not present the problem in this way. Instead, learn mathjax and present the problem accordingly. Second of all, in order to use partial fractions it is best to reduce $frac{x^4}{4-x^2}$ to a proper fraction. This means: divide $x^4$ by $4-x^2.$ The quotient will be easy to integrate and you will then apply partial fractions to the remainder. Third of all, if you are still having problems, show significant work in your query, and elaborate on exactly where you are still having problems.
$endgroup$
– user2661923
Jan 17 at 23:39
add a comment |
$begingroup$
I'm having a lot of trouble with this integral. I don't know what to set A+B equal to. There's an x^4 in the numerator and I'm trying to figure out how to account for it. What am I doing wrong? Thanks.
Problem: $$intfrac{x^4}{4-x^2}space dx$$
What I have tried:
$$frac{x^4}{(2-x)(2+x)}=frac{a}{2-x}+frac{b}{2+x}=frac{a(2+x)+b(2-x)}{(2-x)(2+x)}implies x^4=(a+b)x+2(a+b)$$
integration partial-fractions
edited Jan 18 at 0:02
coreyman317
774420
asked Jan 17 at 23:06
J.W.J.W.
524
$endgroup$
I'm having a lot of trouble with this integral. I don't know what to set A+B equal to. There's an x^4 in the numerator and I'm trying to figure out how to account for it. What am I doing wrong? Thanks.
Problem: $$intfrac{x^4}{4-x^2}space dx$$
What I have tried:
$$frac{x^4}{(2-x)(2+x)}=frac{a}{2-x}+frac{b}{2+x}=frac{a(2+x)+b(2-x)}{(2-x)(2+x)}implies x^4=(a+b)x+2(a+b)$$
integration partial-fractions
integration partial-fractions
edited Jan 18 at 0:02
coreyman317
774420
asked Jan 17 at 23:06
J.W.J.W.
524
edited Jan 18 at 0:02
coreyman317
774420
asked Jan 17 at 23:06
J.W.J.W.
524
edited Jan 18 at 0:02
coreyman317
774420
edited Jan 18 at 0:02
coreyman317
774420
edited Jan 18 at 0:02
coreyman317
774420
774420
asked Jan 17 at 23:06
J.W.J.W.
524
asked Jan 17 at 23:06
J.W.J.W.
524
asked Jan 17 at 23:06
J.W.J.W.
524
524
1
$begingroup$
First of all, do not present the problem in this way. Instead, learn mathjax and present the problem accordingly. Second of all, in order to use partial fractions it is best to reduce $frac{x^4}{4-x^2}$ to a proper fraction. This means: divide $x^4$ by $4-x^2.$ The quotient will be easy to integrate and you will then apply partial fractions to the remainder. Third of all, if you are still having problems, show significant work in your query, and elaborate on exactly where you are still having problems.
$endgroup$
– user2661923
Jan 17 at 23:39
add a comment |
1
$begingroup$
First of all, do not present the problem in this way. Instead, learn mathjax and present the problem accordingly. Second of all, in order to use partial fractions it is best to reduce $frac{x^4}{4-x^2}$ to a proper fraction. This means: divide $x^4$ by $4-x^2.$ The quotient will be easy to integrate and you will then apply partial fractions to the remainder. Third of all, if you are still having problems, show significant work in your query, and elaborate on exactly where you are still having problems.
$endgroup$
– user2661923
Jan 17 at 23:39
1
1
$begingroup$
First of all, do not present the problem in this way. Instead, learn mathjax and present the problem accordingly. Second of all, in order to use partial fractions it is best to reduce $frac{x^4}{4-x^2}$ to a proper fraction. This means: divide $x^4$ by $4-x^2.$ The quotient will be easy to integrate and you will then apply partial fractions to the remainder. Third of all, if you are still having problems, show significant work in your query, and elaborate on exactly where you are still having problems.
$endgroup$
– user2661923
Jan 17 at 23:39
$begingroup$
First of all, do not present the problem in this way. Instead, learn mathjax and present the problem accordingly. Second of all, in order to use partial fractions it is best to reduce $frac{x^4}{4-x^2}$ to a proper fraction. This means: divide $x^4$ by $4-x^2.$ The quotient will be easy to integrate and you will then apply partial fractions to the remainder. Third of all, if you are still having problems, show significant work in your query, and elaborate on exactly where you are still having problems.
$endgroup$
– user2661923
Jan 17 at 23:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When the numerator has higher degree, you should do long division before partial fractions.
$$frac{x^4}{4-x^2} = -x^2 - 4 - frac{16}{x^2-4}.$$
Now you can do partial fractions on the last term.
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
$endgroup$
$begingroup$
Oh wow I knew that. Just escaped me for a second. Thank you.
$endgroup$
– J.W.
Jan 17 at 23:49
add a comment |
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1 Answer
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$begingroup$
When the numerator has higher degree, you should do long division before partial fractions.
$$frac{x^4}{4-x^2} = -x^2 - 4 - frac{16}{x^2-4}.$$
Now you can do partial fractions on the last term.
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
$endgroup$
$begingroup$
Oh wow I knew that. Just escaped me for a second. Thank you.
$endgroup$
– J.W.
Jan 17 at 23:49
add a comment |
$begingroup$
When the numerator has higher degree, you should do long division before partial fractions.
$$frac{x^4}{4-x^2} = -x^2 - 4 - frac{16}{x^2-4}.$$
Now you can do partial fractions on the last term.
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
$endgroup$
$begingroup$
Oh wow I knew that. Just escaped me for a second. Thank you.
$endgroup$
– J.W.
Jan 17 at 23:49
add a comment |
$begingroup$
When the numerator has higher degree, you should do long division before partial fractions.
$$frac{x^4}{4-x^2} = -x^2 - 4 - frac{16}{x^2-4}.$$
Now you can do partial fractions on the last term.
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
$endgroup$
When the numerator has higher degree, you should do long division before partial fractions.
$$frac{x^4}{4-x^2} = -x^2 - 4 - frac{16}{x^2-4}.$$
Now you can do partial fractions on the last term.
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
answered Jan 17 at 23:31
angryavianangryavian
41.7k23381
41.7k23381
$begingroup$
Oh wow I knew that. Just escaped me for a second. Thank you.
$endgroup$
– J.W.
Jan 17 at 23:49
add a comment |
$begingroup$
Oh wow I knew that. Just escaped me for a second. Thank you.
$endgroup$
– J.W.
Jan 17 at 23:49
$begingroup$
Oh wow I knew that. Just escaped me for a second. Thank you.
$endgroup$
– J.W.
Jan 17 at 23:49
$begingroup$
Oh wow I knew that. Just escaped me for a second. Thank you.
$endgroup$
– J.W.
Jan 17 at 23:49
add a comment |
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$begingroup$
First of all, do not present the problem in this way. Instead, learn mathjax and present the problem accordingly. Second of all, in order to use partial fractions it is best to reduce $frac{x^4}{4-x^2}$ to a proper fraction. This means: divide $x^4$ by $4-x^2.$ The quotient will be easy to integrate and you will then apply partial fractions to the remainder. Third of all, if you are still having problems, show significant work in your query, and elaborate on exactly where you are still having problems.
$endgroup$
– user2661923
Jan 17 at 23:39