Question

Find the integral of:

Answer 1

\[\int\limits_{}^{}\frac{e ^{\frac{ 1 }{x }}dx }{ x^{3} }\]

Answer 2

Let u= 1/x, du=1/x^2 ?

Answer 3

Substitute \(1/x = u\) .

Answer 4

\( du = -1/x^2 dx\)

Answer 5

What is the form of integral in terms of u now ?

Answer 6

\[\int\limits_{}^{}\frac{ u du }{ x }\]

Answer 7

Is that right?

Answer 8

With the negative sign.. :)

Answer 9

It should completely be in terms of u. There should be no x. But this is wrong !

Answer 10

\[-\int\limits_{}^{}e^{u}du\]

Answer 11

It should be \(-\int ue^u du\) .

Answer 12

just to share an alternative method,

you can also put \(\Large u = e^{\dfrac{1}{x}}\)
but even then you will have to use product rule.
so you can go ahead with your substitution :)

Answer 13

Now apply integration by parts.

Answer 14

\[-\frac{ u ^{2} }{ 2 } \left( e ^{u} \right)\]

Answer 15

+c

Answer 16

Not sure...

Answer 17

The answer is wrong. Do know integration by parts ?

Answer 18

Nope.. :)

Answer 19

OK. If you have an integral like \(\int u(x)v(x) dx\), you can write it as
\(\int U(x)V(x) dx = U(x) \int V(x) dx - \int \left(\frac{dU(x)}{dx} \int V(x)dx \right) dx\)
Here x is u, U(u) is also u and V(u) = \(e^u\).

Answer 20

Can you apply this ?

Answer 21

Nope.. :( We haven't discuss that yet..

Answer 22

Hmm. Try to work it out.

Answer 23

So,
\(\int u e^u du = u\int e^u du - \int \left(\frac{du}{du} \int e^udu\right)du = ue^u - \int e^udu = e^u(u-1)\)
So \(- \int u e^u du = e^u(1-u)\) + const.