Question

Integration...(q below)

Answer 1

\[\int\limits_{}^{} (1+x-\frac{ 1 }{ x }dx) e ^{x+1/x}\]

Answer 2

any idea?

Answer 3

\[u=x+\frac{1}{x} \\ du=(1-\frac{1}{x^2}) dx \\ \\ 1+x-\frac{1}{x}=1+x(1-\frac{1}{x^2}) ... \text{ hmmm } \\\ \int\limits (1+x(1-\frac{1}{x^2}))e^{x+\frac{1}{x}} dx \\ =\int\limits [ 1 \cdot e^{x+\frac{1}{x}}+x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}) dx \\ \\ \text{ think product rule }\]

Answer 4

it helps you can make those one replacements to make it look just a little more easier to see
\[ x \cdot e^u+x \frac{du}{dx}e^{u} =\]

Answer 5

oops

Answer 6

that one x is suppose to be a 1

Answer 7

\[1 \cdot e^u+x \frac{du}{dx}e^{u} =\]

Answer 8

how to integrate this..wrt what should i integrate?? @myininaya

Answer 9

@freckles can u help?

Answer 10

@zepdrix

Answer 11

Ohhh product rule,
ah yes, I see it now XD
Hah, that's clever

Answer 12

Stuck somewhere in this mess of a problem? :o

Answer 13

Can you state the problem more clearly? I think there is a typo there!

Answer 14

how to continue after that step..?

Answer 15

with respect to what should i integrate this..?

Answer 16

\(=\int\limits [ 1 \cdot e^{x+\frac{1}{x}}+x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}) dx \\ \\ \text{ think product rule } \\ \int f'(x)g(x) + g'(x)f(x) dx\\ where, f(x) = x, \: g(x)= e^{x+\frac{1}{x}} \\ = \int (f(x)g(x))'dx \\ = f(x)g(x) + c \\ = ....\)

Answer 17

oh ok now i understand.. thanks @hartnn

Answer 18

welcome ^_^