Question

Integrals,

http://screencast.com/t/zH6EYVKygWUy

Answer 1

what is the problem?

Answer 2

are you asking how to get from the second to last line to the last line ?

they factored out e^x

they renamed - 2 C1 to C (C1 is an arbitrary constant, and so is -2 * C1 )

Answer 3

So those steps are unnecessary ? My answer is correct too ? @phi @surjithayer

Answer 4

what is the original problem?

it looks like they are integrating \( \int x^2 e^x dx \)
if so, you do it in steps.
the part that you boxed, labeled "I got this far" is the integral \( \int x e^x dx \)
which you need to finish the harder integral \( \int x^2 e^x dx \)

Answer 5

we don't know you are correct or not,only after seeing we can tell.
\[otherwise \int\limits u v dx=u \int\limits v dx-\int\limits u'( \int\limits v dx) dx+c,where u and v are functions of x.\]

Answer 6

http://screencast.com/t/IlgNrFitG

Answer 7

http://screencast.com/t/IlgNrFitG

Answer 8

http://screencast.com/t/yp19CmsJhD

Answer 9

this is all

Answer 10

yes.. as I assumed...
notice after step 1, you have to do the integral \( \int x e^x dx \)
step 2 evaluates that integral

step 3 takes the result of step 2, and uses it to finish step 1

Answer 11

:O first time I am seeing this

Answer 12

Thanks for the answer

Answer 13

does it make sense ?

Answer 14

One last question whats the integral of ln(x) ????

Answer 15

Yes I understood

Answer 16

I use wolfram for those kind of questions

Answer 17

:/

Answer 18

it is x (-1+ln(x)) + C

Answer 19

\[I=\int\limits \ln x*1 dx=\ln x *x-\int\limits \frac{ 1 }{x}*x dx+c=\ln x*x-x+c=x \left( \ln x-1 \right)+c\]

Answer 20

thanks sir !!

Answer 21

yw