Integration work: $intsqrtfrac2-xx-3 mathrm dx$
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$$intsqrtdfrac2-xx-3mathrm dx$$
My approach
I=$$intsqrtdfrac2-xx-3mathrm dx$$
I= $$intfrac2-xsqrt-x^2+5x-6mathrm dx$$
Next I substituted 2-x =t and processed but I am not getting the answer. Can you guys help me with this
calculus integration
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up vote
2
down vote
favorite
$$intsqrtdfrac2-xx-3mathrm dx$$
My approach
I=$$intsqrtdfrac2-xx-3mathrm dx$$
I= $$intfrac2-xsqrt-x^2+5x-6mathrm dx$$
Next I substituted 2-x =t and processed but I am not getting the answer. Can you guys help me with this
calculus integration
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$$intsqrtdfrac2-xx-3mathrm dx$$
My approach
I=$$intsqrtdfrac2-xx-3mathrm dx$$
I= $$intfrac2-xsqrt-x^2+5x-6mathrm dx$$
Next I substituted 2-x =t and processed but I am not getting the answer. Can you guys help me with this
calculus integration
$$intsqrtdfrac2-xx-3mathrm dx$$
My approach
I=$$intsqrtdfrac2-xx-3mathrm dx$$
I= $$intfrac2-xsqrt-x^2+5x-6mathrm dx$$
Next I substituted 2-x =t and processed but I am not getting the answer. Can you guys help me with this
calculus integration
calculus integration
edited Sep 7 at 23:25
amWhy
190k26221433
190k26221433
asked Sep 4 at 12:24
harambe
1135
1135
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add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Hint: try to compute $I$ via the substitution
$$
frac2-xx-3=t
$$
Can I ask you what is the idea behind this substitution?
â harambe
Sep 4 at 13:12
@harambe It is a standard substitution that I learned as a student, I can't provide any extraordinary intuition behind it.
â Siminore
Sep 4 at 14:29
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up vote
2
down vote
Practical way is to choose $$sqrtdfrac2-xx-3=t$$ so, $x=frac3t^2+2t^2+1$ ans then...
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up vote
2
down vote
$$frac2-xx-3=t^2$$
$$frac1(x-3)^2dx=2tdt$$
$$x=frac3t^2+2t^2+1$$
$$x-3=frac3t^2+2t^2+1-3=-frac1t^2+1$$
$$int tcdot2tdtcdot(frac-1t^2+1)^2=intfrac2t^2(t^2+1)^2dt$$
Now solving
$$displaystyleintdfract^2left(t^2+1right)^2,mathrmdt$$
Write it as
$$displaystyleintdfract^2+1-1left(t^2+1right)^2,mathrmdt=displaystyleintdfrac1left(t^2+1right),mathrmdt-displaystyleintdfrac1left(t^2+1right)^2,mathrmdt$$
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: try to compute $I$ via the substitution
$$
frac2-xx-3=t
$$
Can I ask you what is the idea behind this substitution?
â harambe
Sep 4 at 13:12
@harambe It is a standard substitution that I learned as a student, I can't provide any extraordinary intuition behind it.
â Siminore
Sep 4 at 14:29
add a comment |Â
up vote
2
down vote
accepted
Hint: try to compute $I$ via the substitution
$$
frac2-xx-3=t
$$
Can I ask you what is the idea behind this substitution?
â harambe
Sep 4 at 13:12
@harambe It is a standard substitution that I learned as a student, I can't provide any extraordinary intuition behind it.
â Siminore
Sep 4 at 14:29
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: try to compute $I$ via the substitution
$$
frac2-xx-3=t
$$
Hint: try to compute $I$ via the substitution
$$
frac2-xx-3=t
$$
answered Sep 4 at 12:30
Siminore
29.9k23167
29.9k23167
Can I ask you what is the idea behind this substitution?
â harambe
Sep 4 at 13:12
@harambe It is a standard substitution that I learned as a student, I can't provide any extraordinary intuition behind it.
â Siminore
Sep 4 at 14:29
add a comment |Â
Can I ask you what is the idea behind this substitution?
â harambe
Sep 4 at 13:12
@harambe It is a standard substitution that I learned as a student, I can't provide any extraordinary intuition behind it.
â Siminore
Sep 4 at 14:29
Can I ask you what is the idea behind this substitution?
â harambe
Sep 4 at 13:12
Can I ask you what is the idea behind this substitution?
â harambe
Sep 4 at 13:12
@harambe It is a standard substitution that I learned as a student, I can't provide any extraordinary intuition behind it.
â Siminore
Sep 4 at 14:29
@harambe It is a standard substitution that I learned as a student, I can't provide any extraordinary intuition behind it.
â Siminore
Sep 4 at 14:29
add a comment |Â
up vote
2
down vote
Practical way is to choose $$sqrtdfrac2-xx-3=t$$ so, $x=frac3t^2+2t^2+1$ ans then...
add a comment |Â
up vote
2
down vote
Practical way is to choose $$sqrtdfrac2-xx-3=t$$ so, $x=frac3t^2+2t^2+1$ ans then...
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Practical way is to choose $$sqrtdfrac2-xx-3=t$$ so, $x=frac3t^2+2t^2+1$ ans then...
Practical way is to choose $$sqrtdfrac2-xx-3=t$$ so, $x=frac3t^2+2t^2+1$ ans then...
answered Sep 4 at 12:32
mrs
58.3k750143
58.3k750143
add a comment |Â
add a comment |Â
up vote
2
down vote
$$frac2-xx-3=t^2$$
$$frac1(x-3)^2dx=2tdt$$
$$x=frac3t^2+2t^2+1$$
$$x-3=frac3t^2+2t^2+1-3=-frac1t^2+1$$
$$int tcdot2tdtcdot(frac-1t^2+1)^2=intfrac2t^2(t^2+1)^2dt$$
Now solving
$$displaystyleintdfract^2left(t^2+1right)^2,mathrmdt$$
Write it as
$$displaystyleintdfract^2+1-1left(t^2+1right)^2,mathrmdt=displaystyleintdfrac1left(t^2+1right),mathrmdt-displaystyleintdfrac1left(t^2+1right)^2,mathrmdt$$
add a comment |Â
up vote
2
down vote
$$frac2-xx-3=t^2$$
$$frac1(x-3)^2dx=2tdt$$
$$x=frac3t^2+2t^2+1$$
$$x-3=frac3t^2+2t^2+1-3=-frac1t^2+1$$
$$int tcdot2tdtcdot(frac-1t^2+1)^2=intfrac2t^2(t^2+1)^2dt$$
Now solving
$$displaystyleintdfract^2left(t^2+1right)^2,mathrmdt$$
Write it as
$$displaystyleintdfract^2+1-1left(t^2+1right)^2,mathrmdt=displaystyleintdfrac1left(t^2+1right),mathrmdt-displaystyleintdfrac1left(t^2+1right)^2,mathrmdt$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$frac2-xx-3=t^2$$
$$frac1(x-3)^2dx=2tdt$$
$$x=frac3t^2+2t^2+1$$
$$x-3=frac3t^2+2t^2+1-3=-frac1t^2+1$$
$$int tcdot2tdtcdot(frac-1t^2+1)^2=intfrac2t^2(t^2+1)^2dt$$
Now solving
$$displaystyleintdfract^2left(t^2+1right)^2,mathrmdt$$
Write it as
$$displaystyleintdfract^2+1-1left(t^2+1right)^2,mathrmdt=displaystyleintdfrac1left(t^2+1right),mathrmdt-displaystyleintdfrac1left(t^2+1right)^2,mathrmdt$$
$$frac2-xx-3=t^2$$
$$frac1(x-3)^2dx=2tdt$$
$$x=frac3t^2+2t^2+1$$
$$x-3=frac3t^2+2t^2+1-3=-frac1t^2+1$$
$$int tcdot2tdtcdot(frac-1t^2+1)^2=intfrac2t^2(t^2+1)^2dt$$
Now solving
$$displaystyleintdfract^2left(t^2+1right)^2,mathrmdt$$
Write it as
$$displaystyleintdfract^2+1-1left(t^2+1right)^2,mathrmdt=displaystyleintdfrac1left(t^2+1right),mathrmdt-displaystyleintdfrac1left(t^2+1right)^2,mathrmdt$$
edited Sep 4 at 12:47
answered Sep 4 at 12:37
Deepesh Meena
3,8162825
3,8162825
add a comment |Â
add a comment |Â
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