Evaluate $int{frac{e^{2x}} {sqrt{1-e^x}}} dx$
up vote
3
down vote
favorite
Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$
I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?
integration indefinite-integrals
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
asked May 19 '14 at 1:24
user151911
162
add a comment |
up vote
3
down vote
favorite
Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$
I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?
integration indefinite-integrals
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
asked May 19 '14 at 1:24
user151911
162
Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39
1
Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$
I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?
integration indefinite-integrals
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
asked May 19 '14 at 1:24
user151911
162
Evaluate $$displaystyleint{frac{e^{2x}} {sqrt{1-e^x}}} dx.$$
I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?
integration indefinite-integrals
integration indefinite-integrals
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
asked May 19 '14 at 1:24
user151911
162
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
asked May 19 '14 at 1:24
user151911
162
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
edited Nov 24 at 6:17
Martin Sleziak
44.5k7115268
44.5k7115268
asked May 19 '14 at 1:24
user151911
162
asked May 19 '14 at 1:24
user151911
162
asked May 19 '14 at 1:24
user151911
162
162
Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39
1
Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45
add a comment |
Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39
1
Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45
Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39
Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39
1
1
Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45
Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$
answered May 19 '14 at 1:38
user71352
11.3k21025
by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
– user151911
May 19 '14 at 1:43
1
An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
– user71352
May 19 '14 at 1:46
add a comment |
up vote
0
down vote
$$
intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
$$
After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.
answered May 19 '14 at 2:52
Michael Hardy
1
add a comment |
up vote
0
down vote
$$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
$$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
$$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
$$I=-2int(1-u^2)du$$
$$I=2int(u^2-1)du$$
$$I=2int u^2du-2int du$$
$$I=frac{2u^4}{4}-2u$$
$$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$
answered Nov 11 at 21:22
clathratus
2,300322
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$
answered May 19 '14 at 1:38
user71352
11.3k21025
by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
– user151911
May 19 '14 at 1:43
1
An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
– user71352
May 19 '14 at 1:46
add a comment |
up vote
6
down vote
Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$
answered May 19 '14 at 1:38
user71352
11.3k21025
by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
– user151911
May 19 '14 at 1:43
1
An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
– user71352
May 19 '14 at 1:46
add a comment |
up vote
6
down vote
up vote
6
down vote
Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$
answered May 19 '14 at 1:38
user71352
11.3k21025
Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$
answered May 19 '14 at 1:38
user71352
11.3k21025
answered May 19 '14 at 1:38
user71352
11.3k21025
answered May 19 '14 at 1:38
user71352
11.3k21025
answered May 19 '14 at 1:38
user71352
11.3k21025
11.3k21025
by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
– user151911
May 19 '14 at 1:43
1
An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
– user71352
May 19 '14 at 1:46
add a comment |
by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
– user151911
May 19 '14 at 1:43
1
An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
– user71352
May 19 '14 at 1:46
by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
– user151911
May 19 '14 at 1:43
by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2
– user151911
May 19 '14 at 1:43
1
1
An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
– user71352
May 19 '14 at 1:46
An alternative way to solve this using integration by parts is to notice that $frac{e^{2x}}{sqrt{1-e^{x}}}=frac{e^{x}}{sqrt{1-e^{x}}}e^{x}=bigg(frac{d}{dx}(2 sqrt{1-e^{x}})bigg)e^{x}$
– user71352
May 19 '14 at 1:46
add a comment |
up vote
0
down vote
$$
intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
$$
After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.
answered May 19 '14 at 2:52
Michael Hardy
1
add a comment |
up vote
0
down vote
$$
intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
$$
After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.
answered May 19 '14 at 2:52
Michael Hardy
1
add a comment |
up vote
0
down vote
up vote
0
down vote
$$
intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
$$
After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.
answered May 19 '14 at 2:52
Michael Hardy
1
$$
intfrac{e^x}{sqrt{1-e^x}}underbrace{Big(e^x , dxBig)}_{text{HINT}}
$$
After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.
answered May 19 '14 at 2:52
Michael Hardy
1
answered May 19 '14 at 2:52
Michael Hardy
1
answered May 19 '14 at 2:52
Michael Hardy
1
answered May 19 '14 at 2:52
Michael Hardy
1
1
add a comment |
add a comment |
up vote
0
down vote
$$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
$$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
$$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
$$I=-2int(1-u^2)du$$
$$I=2int(u^2-1)du$$
$$I=2int u^2du-2int du$$
$$I=frac{2u^4}{4}-2u$$
$$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$
answered Nov 11 at 21:22
clathratus
2,300322
add a comment |
up vote
0
down vote
$$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
$$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
$$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
$$I=-2int(1-u^2)du$$
$$I=2int(u^2-1)du$$
$$I=2int u^2du-2int du$$
$$I=frac{2u^4}{4}-2u$$
$$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$
answered Nov 11 at 21:22
clathratus
2,300322
add a comment |
up vote
0
down vote
up vote
0
down vote
$$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
$$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
$$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
$$I=-2int(1-u^2)du$$
$$I=2int(u^2-1)du$$
$$I=2int u^2du-2int du$$
$$I=frac{2u^4}{4}-2u$$
$$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$
answered Nov 11 at 21:22
clathratus
2,300322
$$I=intfrac{e^{2x}}{(1-e^x)^{1/2}}dx$$
$$I=intfrac{e^x}{(1-e^x)^{1/2}}e^xdx$$
Substitution: $u^2=1-e^xRightarrow -2udu=e^xdx$.
$$I=intfrac{1-u^2}{(u^2)^{1/2}}(-2)udu$$
$$I=-2int(1-u^2)du$$
$$I=2int(u^2-1)du$$
$$I=2int u^2du-2int du$$
$$I=frac{2u^4}{4}-2u$$
$$I=frac{(1-e^x)^2}{2}-2sqrt{1-e^x}+C$$
answered Nov 11 at 21:22
clathratus
2,300322
answered Nov 11 at 21:22
clathratus
2,300322
answered Nov 11 at 21:22
clathratus
2,300322
answered Nov 11 at 21:22
clathratus
2,300322
2,300322
add a comment |
add a comment |
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Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem?
– Asimov
May 19 '14 at 1:39
1
Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral
– Triatticus
May 19 '14 at 1:45