Substitution for integrals two different cases.
$begingroup$
In my book they make the point that: "The method of substitution cannot be force to work".
They then go on to say that: "there is no substitution that will do much good with the integral $int x(2+x^7)^{frac{1}{5}}dx $ "
However, the integral $int x^6(2+x^7)^{frac{1}{5}}dx $ does apparently yield a substitution for $2+x^7$. Where $u=2+x^7$
Why is this? What distinguishes the two, obviously the order of the exponent $x$ and $x^6$ does something. But I am unsure about what.
calculus integration indefinite-integrals
edited Dec 24 '18 at 0:25
amWhy
1
asked Dec 23 '18 at 23:59
oxodooxodo
32219
$endgroup$
add a comment |
$begingroup$
In my book they make the point that: "The method of substitution cannot be force to work".
They then go on to say that: "there is no substitution that will do much good with the integral $int x(2+x^7)^{frac{1}{5}}dx $ "
However, the integral $int x^6(2+x^7)^{frac{1}{5}}dx $ does apparently yield a substitution for $2+x^7$. Where $u=2+x^7$
Why is this? What distinguishes the two, obviously the order of the exponent $x$ and $x^6$ does something. But I am unsure about what.
calculus integration indefinite-integrals
edited Dec 24 '18 at 0:25
amWhy
1
asked Dec 23 '18 at 23:59
oxodooxodo
32219
$endgroup$
add a comment |
$begingroup$
In my book they make the point that: "The method of substitution cannot be force to work".
They then go on to say that: "there is no substitution that will do much good with the integral $int x(2+x^7)^{frac{1}{5}}dx $ "
However, the integral $int x^6(2+x^7)^{frac{1}{5}}dx $ does apparently yield a substitution for $2+x^7$. Where $u=2+x^7$
Why is this? What distinguishes the two, obviously the order of the exponent $x$ and $x^6$ does something. But I am unsure about what.
calculus integration indefinite-integrals
edited Dec 24 '18 at 0:25
amWhy
1
asked Dec 23 '18 at 23:59
oxodooxodo
32219
$endgroup$
In my book they make the point that: "The method of substitution cannot be force to work".
They then go on to say that: "there is no substitution that will do much good with the integral $int x(2+x^7)^{frac{1}{5}}dx $ "
However, the integral $int x^6(2+x^7)^{frac{1}{5}}dx $ does apparently yield a substitution for $2+x^7$. Where $u=2+x^7$
Why is this? What distinguishes the two, obviously the order of the exponent $x$ and $x^6$ does something. But I am unsure about what.
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Dec 24 '18 at 0:25
amWhy
1
asked Dec 23 '18 at 23:59
oxodooxodo
32219
edited Dec 24 '18 at 0:25
amWhy
1
asked Dec 23 '18 at 23:59
oxodooxodo
32219
edited Dec 24 '18 at 0:25
amWhy
1
edited Dec 24 '18 at 0:25
amWhy
1
edited Dec 24 '18 at 0:25
amWhy
1
1
asked Dec 23 '18 at 23:59
oxodooxodo
32219
asked Dec 23 '18 at 23:59
oxodooxodo
32219
asked Dec 23 '18 at 23:59
oxodooxodo
32219
32219
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
Consider $$z=2+x^7$$
$$dz=7x^6dx\text{or}$$
$$x^6dx=dz/7$$
This means, $$int x^6(2+x^7)^{1/5}dx=intfrac{1}{7} z^{1/5}dz\text{which can easily be calculated to be equal to }frac{5z^{6/5}}{42}.$$
Note that nothing of this sort could happen in the case $x^1$, since it won't come out to be of the form of the differentiation of $2+x^7$.
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
$endgroup$
1
$begingroup$
I see. Thank you!
$endgroup$
– oxodo
Dec 24 '18 at 0:16
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Consider $$z=2+x^7$$
$$dz=7x^6dx\text{or}$$
$$x^6dx=dz/7$$
This means, $$int x^6(2+x^7)^{1/5}dx=intfrac{1}{7} z^{1/5}dz\text{which can easily be calculated to be equal to }frac{5z^{6/5}}{42}.$$
Note that nothing of this sort could happen in the case $x^1$, since it won't come out to be of the form of the differentiation of $2+x^7$.
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
$endgroup$
1
$begingroup$
I see. Thank you!
$endgroup$
– oxodo
Dec 24 '18 at 0:16
add a comment |
$begingroup$
Consider $$z=2+x^7$$
$$dz=7x^6dx\text{or}$$
$$x^6dx=dz/7$$
This means, $$int x^6(2+x^7)^{1/5}dx=intfrac{1}{7} z^{1/5}dz\text{which can easily be calculated to be equal to }frac{5z^{6/5}}{42}.$$
Note that nothing of this sort could happen in the case $x^1$, since it won't come out to be of the form of the differentiation of $2+x^7$.
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
$endgroup$
1
$begingroup$
I see. Thank you!
$endgroup$
– oxodo
Dec 24 '18 at 0:16
add a comment |
$begingroup$
Consider $$z=2+x^7$$
$$dz=7x^6dx\text{or}$$
$$x^6dx=dz/7$$
This means, $$int x^6(2+x^7)^{1/5}dx=intfrac{1}{7} z^{1/5}dz\text{which can easily be calculated to be equal to }frac{5z^{6/5}}{42}.$$
Note that nothing of this sort could happen in the case $x^1$, since it won't come out to be of the form of the differentiation of $2+x^7$.
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
$endgroup$
Consider $$z=2+x^7$$
$$dz=7x^6dx\text{or}$$
$$x^6dx=dz/7$$
This means, $$int x^6(2+x^7)^{1/5}dx=intfrac{1}{7} z^{1/5}dz\text{which can easily be calculated to be equal to }frac{5z^{6/5}}{42}.$$
Note that nothing of this sort could happen in the case $x^1$, since it won't come out to be of the form of the differentiation of $2+x^7$.
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
answered Dec 24 '18 at 0:08
Ankit KumarAnkit Kumar
1,427221
1,427221
1
$begingroup$
I see. Thank you!
$endgroup$
– oxodo
Dec 24 '18 at 0:16
add a comment |
1
$begingroup$
I see. Thank you!
$endgroup$
– oxodo
Dec 24 '18 at 0:16
1
1
$begingroup$
I see. Thank you!
$endgroup$
– oxodo
Dec 24 '18 at 0:16
$begingroup$
I see. Thank you!
$endgroup$
– oxodo
Dec 24 '18 at 0:16
add a comment |
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