Question

Solve the indefinite integral:
e^(2x) / x^3

Answer 1

where is the dx??

Answer 2

on way might be to write it as e^(2x) x^(-3) and use integration by parts

Answer 3

But when using integration by parts for that, wouldn't it just go on forever, nesting integration by parts?

Answer 4

yup,
and make
u = x^-3
dv = e^2x

Answer 5

no, it will be just three IBP

Answer 6

Yep, what Jimmy said works, just do it reverse from what you usually do. Thanks Jimmy!

Answer 7

but if u=x^(-3), then every time you differentiate u, the exponent gets farther from 0. So when do the IBPs stop?

Answer 8

good point, i didnt think about that.

:-)

Answer 9

I think you are talaking about the tabular method in IBP , try using the direct theorem

Answer 10

how?

Answer 11

What's the direct theorem?

Answer 12

is it:
uv-integral of(v(du))???

Answer 13

But then again, the damn du/dx is 2, not 2x obviously. So you get pellet from IBP.
And yes, you just add up on using u=x^-3...
Direct theorem?

Answer 14

yes. it is. @kumarsajan

Answer 15

but when you use that it will make the du infinite if you chose x^-3 as your u,
and if it is e^2x it is also infinite.

Answer 16

I see !

Answer 17

But, by using u=e^2x and dv=x^-3, you actually reduce x^-3 by integrating. So it eventually solves itself.

Answer 18

thats the way: - u = e^2x

Answer 19

i see then, as x becomes 0 it will be solved.

:-)

Answer 20

that's right