Question

in integrating lnx dx why is the v=x?

Answer 1

v=x?

Answer 2

how do i solve \[\int\limits_{?}^{?} lnx dx\]

Answer 3

you need to do integration by parts

Answer 4

by parts sorry

Answer 5

u know integral by parts?

Answer 6

using 1 dx and lnx as your dv and u

Answer 7

yes. i know integration by parts. but how to know when we use integration by parts?

Answer 8

when a product of functions is ther or when u dont see any other method..when the integral is not a known form

Answer 9

usually i resort to integration by parts when there is no other easier method to solve the integral such as substitution or partial fractions

Answer 10

because when u integrate by parts
yu need u and dv
and for u ...
the priority is for ( logarithms then inverses then algebraic then trigonometric then exponentials)
so its 1*lnx dx
so for u
you choose lnx since logarithms hav the highest priority
and for dv you choos 1* dx
then u find du and v
so integrating dv=dx
u get v=x

u use by parts when u hav two fxns multiplied by eachother

Answer 11

i see. and when do we use substitution?

Answer 12

subsitution is usually the first thing i try out since it is easy to test and look for

Answer 13

u can use anything as long as it simplifies ur integral..that knowledge only comes with practice

Answer 14

also check if you can break up the numerator

Answer 15

ok. another question sorry.: why is the integral of e^-x is -e^-x?

Answer 16

why is it negative?

Answer 17

use substitution

Answer 18

because
integration of e^f(x) = e^f(x)/ f '(x)

Answer 19

with u = -x then du = -1

Answer 20

du=-1 dx*

Answer 21

and remember that d/dx e^x = e^x then its antiderivative is also true that \[\int\limits_{}^{}e^x =e^x\]

Answer 22

if you take d/dx -e^-x you will see that it will become e^-x

Answer 23

ok. how do I solve this?..sorry got a test tomorrow need to have great score.\[\int\limits_{0}^{1} \arctan x dx\] :

Answer 24

the same thing that you do to solve for \[\int\limits_{}^{}lnx dx\]

Answer 25

still integration by parts?

Answer 26

(im ignoring the limits of integration since its simple plug in numbers) but integration by parts since it isnt a antiderivative of a easily known trig function

Answer 27

so you take u = tanx and dv = dx

Answer 28

you can also imagine it as \[\int\limits_{}^{}arctanx (1) dx\]

Answer 29

why is the u only tanx? why not arctan x?

Answer 30

and becomes du = csc^2 and so forth

Answer 31

o whoops i mean arctan

Answer 32

=)

Answer 33

I am confused with \[\int\limits_{?}^{?} e^x cosx dx\]. I already started solving but i'm stuck with \[\int\limits_{?}^{?} sinx e^x dx\]

Answer 34

this one is a special case of integration by parts

Answer 35

do you know how to use the table method for integration by parts?

Answer 36

well i call it the table method i think it has a name

Answer 37

table method? no. i never heard of that.

Answer 38

what is it?

Answer 39

you make 2 columns, one for u and one for dv

Answer 40

hmm, it takes a bit of explainning

Answer 41

http://www.hyper-ad.com/tutoring/int_parts.htm shows a use of it

Answer 42

thanks. really. one more thing do u know where i can get examples for integration using power substitution? i didn't get a lot of examples from my search.

Answer 43

well i dont think its nesscary to solve by parts but it makes it faster if you need to do by parts multiple times

Answer 44

power substitution?

Answer 45

yes. my teacher called it as power substitution. it has square roots.

Answer 46

from your earlier problem of \[\int\limits_{ }^{ }e^xcosx dx\]you need to do by parts twice and then add the similar terms together and then divide by two

Answer 47

so i need to do another by part with\[\int\limits_{?}^{?} sinx e^x dx?\]

Answer 48

well square roots are the same thing as powers like x(2x^2-3x)^(1/2) can be easily seen through with substitution

Answer 49

yep

Answer 50

you should get \[\int\limits_{ }^{} sinx e^x dx = uv - \int\limits_{}^{}vdu\]

Answer 51

how do you know what to substitute?

Answer 52

but when you take \[\int\limits_{ }^{ } v du\] you get an addition by parts with \[- \int\limits_{}^{}sinx e^x dx\] which you use to add to both sides and divide by 2

Answer 53

if you see a grouped function and its derivative outside that grouped function you can use substitution

Answer 54

so like x(x^2-3) you can take a sub with the x^2-3

Answer 55

constants can be easily adjusted to fit the sub if nessecary

Answer 56

so you would get du = 2x but theres only a x
but you can also write x as 2x/2

Answer 57

in my earlier problem: i came up again with e^xsinx+ e^x cosx + \[\int\limits\limits_{?}^{?} cosx e^x dx\] and i was stuck with the cosx e^x dx

Answer 58

can u go here: http://www.twiddla.com/565703

Answer 59

yes.

Answer 60

i finished the last part and that should be the final answer for that problem http://www.twiddla.com/565703