Question

find integral of 1/x(logx)^m dx,m>o




does this question have 2 answers?

Answer 1

why 2 answers?

Answer 2

m=1 and m=2toinfinity

Answer 3

let \[u = \log x\]
\[du = \frac 1x dx\]
does that help?

Answer 4

and no that's not a limit... m>0 is just to denote m exists

Answer 5

your answer should be in algebra form

Answer 6

ya i know that but when m=1 its logu,otherwise u need to use the other law...x^n...

Answer 7

ok

Answer 8

oh well then yes you ahve 2 answers

Answer 9

Do this............
find the integral of 1/x^3-x^4

Answer 10

ok

Answer 11

\[\int \frac{1}{x^3 } - x^4\]
???

Answer 12

no x4 is in the denominator along wid x^3

Answer 13

the one which u thought is damn easy

Answer 14

x^4

Answer 15

\[\huge \int \frac 1{x^2 - x^4} \implies \int \frac 1{x^3(1 - x)}\]
does that help?

Answer 16

yes
then...

Answer 17

uhh that should be 1/(x^3 - x^4)

you know what i mean...

Answer 18

yes..

Answer 19

sry,..

Answer 20

now you use partil fractions...

\[\Large \implies \frac Ax + \frac B{x^2} + \frac C{x^3} + \frac D{1-x} = \frac{1}{x^3(1-x)}\]
right?

Answer 21

i did it some other way...lol
maybe thats also correct
..that is defintely correct

Answer 22

what did you do?

Answer 23

integration by parts?

Answer 24

i put the N=1=(1-x)+x,then split ..then PFM

Answer 25

partial fraction(PFM)

Answer 26

hmm i guess if it works then it's the same

Answer 27

yes

Answer 28

im not familiar with that kind of manipulation

Answer 29

now this one...find integral of x*whole root of 1-x/1+x

Answer 30

\[\int x \sqrt{\frac{1-x}{1+x}}\]
???

Answer 31

yes

Answer 32

how do u type like that?

Answer 33

i have no idea

Answer 34

the integral i mean

Answer 35

k
i tried then...i multiplied N and D by the N...

Answer 36

as for my typing...do you see that Equation button below the reply box?

Answer 37

nah!

Answer 38

ok yes i can see..

Answer 39

i think what you did is right

Answer 40

r u a professor or anythng like that?

Answer 41

yes. something like that

Answer 42

anyway...if you multiply N and D by N you get
\[\int \frac{x(1-x)}{\sqrt{1-x^2}}\]
right?

Answer 43

but then what?
shud i put N=A(d/dxD)+B?

Answer 44

yes

Answer 45

split it up now
\[\int \frac{x}{\sqrt{1-x^2}} + \int \frac{x^2}{\sqrt{1-x^2}}\]
first one is solveable by u-sub

second one is solveable by trig sub

Answer 46

right?

Answer 47

oh ya!!!!!!111111!

Answer 48

yes

Answer 49

btw did u solve that question.............-->find integral of 1+cosxcosA/cosA+cosx ?

Answer 50

where A is constant

Answer 51

\[\int \frac{1+\cos x \cos A}{\cos A + \cos x}\]
???

Answer 52

yes

Answer 53

...well you can try splittig it up again...

Answer 54

what i did was i put 1=sin^2x+cos^2x

Answer 55

took cosx common
and split

Answer 56

hmm clever

Answer 57

but then it becomes s^2x/coA+cosx

Answer 58

s means sine

Answer 59

this is a hard one

Answer 60

ya....these were the ones i didnt understand from the 400 sums

Answer 61

sorry. i have no idea what this is

Answer 62

np

Answer 63

good questions though

Answer 64

ya
........................HOW to do this question here







integrate (3sinx-4)cosx/(5sinx+3)^8 w.r.t.x

Answer 65

wait

Answer 66

there is dx after w.r.t.x

Answer 67

hmm ho many of these do you have

Answer 68

before*

Answer 69

this might be the lst one

Answer 70

last*

Answer 71

let u = sinx
du = cosx dx
\[\implies \int \frac{(3u-4)du}{(5u+3)^8}\]

Answer 72

yes

Answer 73

ok

Answer 74

now by PFM?

Answer 75

:O

Answer 76

no... i think you should just split the fraction

Answer 77

\[\implies \int\frac{3u}{(5u+3)^8} - \int \frac{4}{(5u+3)^8}\]
maybe?

Answer 78

ugh my brain hurts now o.O

Answer 79

lol