Question

integral of (x-1)e^x/x^2?

Answer 1

1) distribute
2) split fraction into 2 integrals
3) solve first integral
4) use by parts on second integral with dv=1/x^2
5) use by parts again with dv=1/x

Answer 2

integral of e^x/x?

Answer 3

yup... by parts again... with dv=1/x

Answer 4

no but for the first integral, not the second. Do I use parts 3 times?

Answer 5

\[\int\limits_{}^{}\frac{1}{x}dx-\int\limits_{}^{}\frac{e ^{x}}{x ^{2}}dx\]

Answer 6

isn't the first one supposed to be integral of e^x/x, not 1/x?

Answer 7

\[\ln \left| x \right|+\frac{2e ^{x}}{x}+\int\limits_{}^{}\frac{e ^{x}}{x}dx\]

Answer 8

are you looking at\[\int\limits_{}^{}(x-1)\frac{e ^{x}}{x ^{2}}dx\]?

Answer 9

oops... yes... by parts on the first one too then

Answer 10

yes, which is integral of xe^x/x^2 - integral of e^x/x^2

Answer 11

u = e^x or does u =1/x?

Answer 12

u=e^x

Answer 13

that's what i just did and it looks like im in a loop.

Answer 14

yuck... I agree

Answer 15

ill try making 1/x = u

Answer 16

If you do the other by parts twice... I think that you might get something from it though.

Answer 17

Sorry.. I'm thinking backward:
\[\frac{d}{dx}(\frac{e^x}{x}) = \frac{xe^x - e^x}{x^2}\]So....

Answer 18

another loop???

Answer 19

\[\int\limits_{ }^{ } e ^{x} *\frac{ 1 }{ x} = \frac{ e ^{x} }{ x } -\int\limits_{ }^{ }-e ^{x} *\frac{ 1 }{ x ^{2} }\]

Answer 20

add integral e^x * -1/x^2 to both sides

Answer 21

yes... I am getting one too.... working.

Answer 22

there we go.. it's recursive

Answer 23

got that, but when you do parts again on that answer algebraic! i get another loop since the new integral is integral of vdu = 2*integeral of e^x/x^3

Answer 24

naw you're done.

Answer 25

think about it for a bit.

Answer 26

PM me when 'eureka'

Answer 27

oh wait so its just e^x/x?

Answer 28

:)

Answer 29

haha wow i kept trying to integrate but the integrals are opposites of each other. Thank you so much.

Answer 30

glad to be of modest service:)

Answer 31

awww man.... nice call