Question

How about this.. Integration

Answer 1

split it into two terms

Answer 2

you mean.. 2/e^x and e^x/e^x

Answer 3

exactly!

Answer 4

it would be 2e^-x + 1 +C?

Answer 5

\[\large \int \dfrac{2+e^x}{e^x}~dx\]

Answer 6

\[\large \int \dfrac{2}{e^x}+\dfrac{e^x}{e^x} ~dx\]

Answer 7

\[\large \int 2e^{-x}+1~dx\]

Answer 8

integrate now..

Answer 9

okay =) wait

Answer 10

i got 2e^-x +x +c

Answer 11

small mistake,
whats the integral of \(\large \int e^{-x} dx\) ?

Answer 12

-e^x

Answer 13

so.. ?

Answer 14

its -2e^x +x +c

Answer 15

Yes !

Answer 16

good job :)

Answer 17

Thank you so much!!

Answer 18

wait @ganeshie8

Answer 19

yes

Answer 20

how do i solve this one?

Answer 21

hey wait a sec, for the previous problem, you got
its -2e^(-x) +x +c

right ?

Answer 22

-2e^x +x

Answer 23

nope, and who ate the C ?

Answer 24

\[\large \int 2e^{-x}+1~dx\]

\[\large - 2e^{-x}+x + C\]

Answer 25

ah. missed the C. hahaha

Answer 26

thats the final answer ^^

Answer 27

=)

Answer 28

for the next problem, substitute \(\large u = \dfrac{\pi}{x}\)

Answer 29

so the derivative would be -pi/x^2?

Answer 30

yes!

Answer 31

and im lost hahaha

Answer 32

\[\int \dfrac{1}{x^2} \sin \left(\frac{\pi}{x}\right) ~dx\]

Answer 33

what do you get after making the substitution ?

Answer 34

really have no idea..

Answer 35

sub \(\large u = \frac{\pi}{x} \)

\(\large du = -\frac{\pi}{x^2}dx\)

\(\large -\frac{du}{\pi} = \frac{1}{x^2}dx\)

Answer 36

substitute these

Answer 37

\[\int \sin \left(u \right) ~\left(-\frac{du}{\pi}\right)\]

Answer 38

\[-\frac{1}{\pi}\int \sin \left(u \right) ~du \]

Answer 39

okay, im following..

Answer 40

goo, integrate ^

Answer 41

that would be -1/pi cos pi/x +c?

Answer 42

whats the integral of sinx ?

Answer 43

its cosx

Answer 44

really sure ?

Answer 45

-cosx

Answer 46

+c

Answer 47

yes ! so whats the final answer

Answer 48

1/pi - cos pi/x +c

Answer 49

@ganeshie8 i need to go. its already 12am here. i need to refresh everything. Thank you so much!! =)) Goodnight

Answer 50

\[-\frac{1}{\pi}\int \sin \left(u \right) ~du\]

\[-\frac{1}{\pi}( - \cos \left(u \right) ) + C\]

\[ \frac{1}{\pi} \cos \left(u \right) + C\]

Answer 51

good luck !!

Answer 52

and have good sleep :)