Evaluate $int frac{7x^4+2}{x^8-e^{7x}}dx$
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Evaluate $$I=int frac{7x^4+2}{x^8-e^{7x}}dx$$
My try: we have $$I=int e^{-7x} times frac{7x^4+2}{x^8e^{-7x}-1}dx$$
I tried to use substitution $xe^{-x}=t$ but of no use.
any clue?
calculus algebra-precalculus indefinite-integrals
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
add a comment |
up vote
1
down vote
favorite
Evaluate $$I=int frac{7x^4+2}{x^8-e^{7x}}dx$$
My try: we have $$I=int e^{-7x} times frac{7x^4+2}{x^8e^{-7x}-1}dx$$
I tried to use substitution $xe^{-x}=t$ but of no use.
any clue?
calculus algebra-precalculus indefinite-integrals
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Evaluate $$I=int frac{7x^4+2}{x^8-e^{7x}}dx$$
My try: we have $$I=int e^{-7x} times frac{7x^4+2}{x^8e^{-7x}-1}dx$$
I tried to use substitution $xe^{-x}=t$ but of no use.
any clue?
calculus algebra-precalculus indefinite-integrals
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
Evaluate $$I=int frac{7x^4+2}{x^8-e^{7x}}dx$$
My try: we have $$I=int e^{-7x} times frac{7x^4+2}{x^8e^{-7x}-1}dx$$
I tried to use substitution $xe^{-x}=t$ but of no use.
any clue?
calculus algebra-precalculus indefinite-integrals
calculus algebra-precalculus indefinite-integrals
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
asked Nov 21 at 11:52
Ekaveera Kumar Sharma
5,50511327
5,50511327
add a comment |
add a comment |
1 Answer
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1
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I don't think it has a closed form. A series form solution to a more general integral in terms of upper incomplete gamma functions is:
$$int frac {at^4+2b}{t^8-be^t} dt = sum_{k=0}^∞ frac {1}{b^k(k+1)^{8k+1}}(frac {a}{b}frac {Gamma [8k+5, (k+1)t]}{(k+1)^4} + 2Gamma [8k+1,(k+1)t])$$
answered Nov 21 at 14:06
Awe Kumar Jha
3189
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I don't think it has a closed form. A series form solution to a more general integral in terms of upper incomplete gamma functions is:
$$int frac {at^4+2b}{t^8-be^t} dt = sum_{k=0}^∞ frac {1}{b^k(k+1)^{8k+1}}(frac {a}{b}frac {Gamma [8k+5, (k+1)t]}{(k+1)^4} + 2Gamma [8k+1,(k+1)t])$$
answered Nov 21 at 14:06
Awe Kumar Jha
3189
add a comment |
up vote
1
down vote
I don't think it has a closed form. A series form solution to a more general integral in terms of upper incomplete gamma functions is:
$$int frac {at^4+2b}{t^8-be^t} dt = sum_{k=0}^∞ frac {1}{b^k(k+1)^{8k+1}}(frac {a}{b}frac {Gamma [8k+5, (k+1)t]}{(k+1)^4} + 2Gamma [8k+1,(k+1)t])$$
answered Nov 21 at 14:06
Awe Kumar Jha
3189
add a comment |
up vote
1
down vote
up vote
1
down vote
I don't think it has a closed form. A series form solution to a more general integral in terms of upper incomplete gamma functions is:
$$int frac {at^4+2b}{t^8-be^t} dt = sum_{k=0}^∞ frac {1}{b^k(k+1)^{8k+1}}(frac {a}{b}frac {Gamma [8k+5, (k+1)t]}{(k+1)^4} + 2Gamma [8k+1,(k+1)t])$$
answered Nov 21 at 14:06
Awe Kumar Jha
3189
I don't think it has a closed form. A series form solution to a more general integral in terms of upper incomplete gamma functions is:
$$int frac {at^4+2b}{t^8-be^t} dt = sum_{k=0}^∞ frac {1}{b^k(k+1)^{8k+1}}(frac {a}{b}frac {Gamma [8k+5, (k+1)t]}{(k+1)^4} + 2Gamma [8k+1,(k+1)t])$$
answered Nov 21 at 14:06
Awe Kumar Jha
3189
answered Nov 21 at 14:06
Awe Kumar Jha
3189
answered Nov 21 at 14:06
Awe Kumar Jha
3189
answered Nov 21 at 14:06
Awe Kumar Jha
3189
3189
add a comment |
add a comment |
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