Mixing
Integration of rational function with substitution
$begingroup$
I am having trouble with integrating this function with substitution.
$
intfrac{x^2-4x}{x-2sqrt{x}}dx
$
First i do this:
$
u=sqrt{x}
$
$
du=frac{1}{2sqrt{x}}dx
$
$
dx=frac{du}{frac{1}{{2sqrt{x}}}}
$
then this:
$
intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
$
But now i don't know if what i have done is correct and know how to proceed.
Any help would be appreciated.
integration
asked Feb 1 at 20:13
krnekikrneki
82
$endgroup$
add a comment |
$begingroup$
I am having trouble with integrating this function with substitution.
$
intfrac{x^2-4x}{x-2sqrt{x}}dx
$
First i do this:
$
u=sqrt{x}
$
$
du=frac{1}{2sqrt{x}}dx
$
$
dx=frac{du}{frac{1}{{2sqrt{x}}}}
$
then this:
$
intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
$
But now i don't know if what i have done is correct and know how to proceed.
Any help would be appreciated.
integration
asked Feb 1 at 20:13
krnekikrneki
82
$endgroup$
add a comment |
$begingroup$
I am having trouble with integrating this function with substitution.
$
intfrac{x^2-4x}{x-2sqrt{x}}dx
$
First i do this:
$
u=sqrt{x}
$
$
du=frac{1}{2sqrt{x}}dx
$
$
dx=frac{du}{frac{1}{{2sqrt{x}}}}
$
then this:
$
intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
$
But now i don't know if what i have done is correct and know how to proceed.
Any help would be appreciated.
integration
asked Feb 1 at 20:13
krnekikrneki
82
$endgroup$
I am having trouble with integrating this function with substitution.
$
intfrac{x^2-4x}{x-2sqrt{x}}dx
$
First i do this:
$
u=sqrt{x}
$
$
du=frac{1}{2sqrt{x}}dx
$
$
dx=frac{du}{frac{1}{{2sqrt{x}}}}
$
then this:
$
intfrac{x^2-4x}{x-2u} frac{du}{frac{1}{{2sqrt{x}}}}
$
But now i don't know if what i have done is correct and know how to proceed.
Any help would be appreciated.
integration
integration
asked Feb 1 at 20:13
krnekikrneki
82
asked Feb 1 at 20:13
krnekikrneki
82
asked Feb 1 at 20:13
krnekikrneki
82
asked Feb 1 at 20:13
krnekikrneki
82
asked Feb 1 at 20:13
krnekikrneki
82
82
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
add a comment |
$begingroup$
If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get
$int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$
for $u=sqrt{x}$ from which you should be able to continue.
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
add a comment |
$begingroup$
Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
add a comment |
$begingroup$
Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
Note that $$(x-2sqrt{x})(x+2sqrt{x})=x^2-4x$$
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Feb 1 at 20:15
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
78.9k42867
add a comment |
add a comment |
$begingroup$
If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get
$int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$
for $u=sqrt{x}$ from which you should be able to continue.
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
$endgroup$
add a comment |
$begingroup$
If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get
$int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$
for $u=sqrt{x}$ from which you should be able to continue.
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
$endgroup$
add a comment |
$begingroup$
If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get
$int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$
for $u=sqrt{x}$ from which you should be able to continue.
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
$endgroup$
If you note that with your substitution follows $x=u^2$ and $x^2=u^4$ you get
$int frac{x^2-4x}{x-2sqrt{x}}dx=2int u^2(u+2)du$
for $u=sqrt{x}$ from which you should be able to continue.
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
answered Feb 1 at 20:54
babemcnuggetsbabemcnuggets
116110
116110
add a comment |
add a comment |
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