Question

i want to confirm the answer of... integrate lnx/x dx

Answer 1

my answer is ln lxl +c

Answer 2

let u=lnx
du=dx/x
\[\int\limits_{}^{}u du=\frac{u^2}{2}+C=\frac{(lnx)^2}{2}+C\]
we can even check this:
\[(\frac{(lnx)^2}{2}+C)'=2*\frac{lnx}{2}*\frac{1}{x}+0=\frac{lnx}{x}\]

Answer 3

do you got it?

Answer 4

yes

Answer 5

k :)

Answer 6

q. how do i know what to choose?

Answer 7

i was looking for \[\int\limits_{}^{}f(x)*f'(x)dx\]

Answer 8

sometimes its not always that easy though

Answer 9

someone said it is like an art and i agree

Answer 10

jajaj..ok

Answer 11

it is not always easy to know what substititon to make but since it was in the form of above it was easy
let u=f(x)
du=f'(x) dx

Answer 12

\[\int\limits_{}^{}\frac{lnx}{x}dx=\int\limits_{}^{}lnx*\frac{1}{x} dx\]

Answer 13

so it was easy for me to notice pretty fast that (lnx)'=1/x
so thats why i let u=lnx

Answer 14

ok..perfect

Answer 15

so how can i integrate tanx dx..... even though i know thats it is equal to -lnlcosxl+c

Answer 16

\[\int\limits_{}^{}\frac{sinx}{cosx} dx\]
right?

Answer 17

tanx=sinx/cosx agree?

Answer 18

ohhhhhhhh... ok

Answer 19

yes

Answer 20

so that du/u

Answer 21

now what do you think the substitution will be?

Answer 22

with the negative sign of the sin

Answer 23

u=cos

Answer 24

du=sin x

Answer 25

sorry
du=-sinx

Answer 26

du=-sinx dx (i like not to forget about this dx part)

Answer 27

oh yea

Answer 28

so show me what that gives us don't integrate yet

Answer 29

then u have
integral of du/u

Answer 30

negative or positive?

Answer 31

- integral of du/u

Answer 32

yes and now integrate

Answer 33

so my result will be
-ln/u/+c

Answer 34

-ln/cosx/+c

Answer 35

yes :)

Answer 36

yeaaa
thank uuuu

Answer 37

np

Answer 38

so whats this \[\int\limits_{}^{}-tanx dx\] equal to?

Answer 39

ln/cosx/+c

Answer 40

becuase the negative sign will change to postive once u take the derivative of cosx

Answer 41

why u pic u =sinx?

Answer 42

accident

Answer 43

u=cosx
du=-sinx dx

Answer 44

:)

Answer 45

good....:

Answer 46

you want to try one i made up? its not hard

Answer 47

ok

Answer 48

\[\int\limits_{}^{}\frac{x+1}{x^2+2x}dx\]

Answer 49

ok.. let me try

Answer 50

u=x^2 + 2x
du=2x+2 dx
=2(x+1) dx
so,
1/2du=x+1 dx

Answer 51

very good so far

Answer 52

1/2 integral of du/u
=1/2 lnlul+c
=1/2 ln/x^2+2xl+c

Answer 53

:)

Answer 54

yeaaaaaaaaaaaaa

Answer 55

i love this stuff

Answer 56

lol it is nice

Answer 57

\[\int\limits_{}^{}\frac{f'(x)}{f(x)}dx\]
so when we have this what will out answer be in the form of?

Answer 58

f(x)*f'(x)

Answer 59

look at the one you just did
didn't we have f'/f dx?

Answer 60

ohh... sorry.. yes yes

Answer 61

=ln|f(x)|+C

Answer 62

yes

Answer 63

\[\int\limits_{}^{}[f(x)]^n*f'(x)dx\]
\[n \neq -1\]
what about this?
what will our answer be?

Answer 64

i will give you a hint
let u=f(x)
so du=f'(x) dx

Answer 65

jjajjjaj... nice hint

Answer 66

where are you from?
you say jajaja alot lol

Answer 67

dominican republic

Answer 68

i am just laughing

Answer 69

jajaaja= laughing?

Answer 70

yess... how can i laugh??

Answer 71

its fine the way you type your laughter
no worries
i like it

Answer 72

lol

Answer 73

so this all makes a little more sense?

Answer 74

the integral stuff?

Answer 75

why?

Answer 76

im asking do you get the integral stuff better?

Answer 77

ohhhhhh,, yess

Answer 78

good :) im fixing to go to sleep goodnight

Answer 79

for me is mornig.. but good nigth,,,

Answer 80

good morning*
i didn't sleep all night lol

Answer 81

jajajajaaja

Answer 82

ayayay.. no good

Answer 83

thank u very much...

Answer 84

np
if you want to try another
this is a good one
\[\int\limits_{}^{}x*\sqrt{x+1} dx\]
the answer is given above in one of the other threads if you want to check yourself

Answer 85

later

Answer 86

ok,, thanks