How to evaluate the integral without using integration by parts
How can we evaluate the following integral by using substitution rule only?
$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$
calculus integration algebra-precalculus
edited Nov 21 at 21:42
M. Strochyk
7,67711119
asked Nov 21 at 21:04
Ninja
1,097720
add a comment |
How can we evaluate the following integral by using substitution rule only?
$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$
calculus integration algebra-precalculus
edited Nov 21 at 21:42
M. Strochyk
7,67711119
asked Nov 21 at 21:04
Ninja
1,097720
What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11
add a comment |
How can we evaluate the following integral by using substitution rule only?
$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$
calculus integration algebra-precalculus
edited Nov 21 at 21:42
M. Strochyk
7,67711119
asked Nov 21 at 21:04
Ninja
1,097720
How can we evaluate the following integral by using substitution rule only?
$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$
calculus integration algebra-precalculus
calculus integration algebra-precalculus
edited Nov 21 at 21:42
M. Strochyk
7,67711119
asked Nov 21 at 21:04
Ninja
1,097720
edited Nov 21 at 21:42
M. Strochyk
7,67711119
asked Nov 21 at 21:04
Ninja
1,097720
edited Nov 21 at 21:42
M. Strochyk
7,67711119
edited Nov 21 at 21:42
M. Strochyk
7,67711119
edited Nov 21 at 21:42
M. Strochyk
7,67711119
7,67711119
asked Nov 21 at 21:04
Ninja
1,097720
asked Nov 21 at 21:04
Ninja
1,097720
asked Nov 21 at 21:04
Ninja
1,097720
1,097720
What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11
add a comment |
What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11
What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11
What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11
add a comment |
1 Answer
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Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
$x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
$$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
therefore,
$$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$
answered Nov 21 at 21:41
M. Strochyk
7,67711119
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
$x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
$$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
therefore,
$$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$
answered Nov 21 at 21:41
M. Strochyk
7,67711119
add a comment |
Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
$x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
$$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
therefore,
$$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$
answered Nov 21 at 21:41
M. Strochyk
7,67711119
add a comment |
Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
$x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
$$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
therefore,
$$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$
answered Nov 21 at 21:41
M. Strochyk
7,67711119
Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
$x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
$$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
therefore,
$$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$
answered Nov 21 at 21:41
M. Strochyk
7,67711119
answered Nov 21 at 21:41
M. Strochyk
7,67711119
answered Nov 21 at 21:41
M. Strochyk
7,67711119
answered Nov 21 at 21:41
M. Strochyk
7,67711119
7,67711119
add a comment |
add a comment |
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What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11