Question

a practice integral.

Answer 1

\[\int\limits_{ }^{ } (\ln x)^2~dx\]

Answer 2

I am assuming by parts.

Answer 3

ya INTEGRAL

Answer 4

Yeah, you are right

Answer 5

By parts.

Answer 6

@cambrige to the rescue

Answer 7

Okay, no one is telling me no, I will proceed.
\[\int\limits_{ }^{ } (\ln x)^2~dx=x(\ln x)^2 -2\int\limits_{ }^{ }(\ln x)~dx\](The Xs cancel)

Answer 8

DUDE.........U KNOW IT.!! -_-

Answer 9

not really a problem then...LOL

Answer 10

\[\int\limits_{ }^{ } (\ln x)^2~dx=x(\ln x)^2 -2\left(\begin{matrix} x (\ln x)-x+C\\ \end{matrix}\right)\]

Answer 11

\[\int\limits_{ }^{ } (\ln x)^2~dx=x(\ln x)^2 -2x (\ln x)+2x+C\]

Answer 12

try this \[\tan^{-1}(lnx)\]

Answer 13

I looked it was un-do-able, but I managed:)

Answer 14

frankly I dunno if it can be solved also

Answer 15

\[\int\limits_{ }^{ } \tan^{-1}\ln(x)~dx\]

Answer 16

what is up with your inverse TANGENTS, LOL?

Answer 17

ya

Answer 18

I love Trig funtions

Answer 19

to give it a better notation.....\[\int\limits_{ }^{ } \tan^{-1}(\ln x)~dx\]wait, you are tricking me again?

Answer 20

I told ya

Answer 21

its same lol

Answer 22

I will take the derivative of tan^-1 (ln x) and do by parts

Answer 23

right?

Answer 24

lol

Answer 25

so by parts would do,
\[\int\limits_{ }^{ } \tan^{-1}(\ln x)~dx=x \tan^{-1}(\ln x)~-\int\limits_{ }^{ }\frac{1}{(\ln x)^2+1}~dx\](X's canceled)

Answer 26

u cant solve it LMAO

Answer 27

I cudnt

Answer 28

if you can't, maybe I still could:P

Answer 29

U m8...try it..

Answer 30

I only learn u-sub in class what do you want? :)

Answer 31

nothing...calm down

Answer 32

think I got it

Answer 33

there must be some sort of an answer to this problem.
using error function (that I don't really know), or a series....

Answer 34

@fbi2015 to which problem?

Answer 35

this

Answer 36

ok that last integral

Answer 37

\[\int\limits_{}^{} \tan^{-1}(lnx) dx\]

Answer 38

I am to the tangent problem, the one I did first is fine

Answer 39

(ln x)^2 I solved:)

Answer 40

I did by parts and stopped at:
\[\int\limits_{ }^{ } \tan^{-1}(\ln x)~dx=x \tan^{-1}(\ln x)~-\int\limits_{ }^{ }\frac{1}{(\ln x)^2+1}~dx \]

Answer 41

you did it correctly, the last integral cannot be solved using elementary functions

Answer 42

but how can it?

Answer 43

It's not a series.

Answer 44

yes, it's "not" it's not an imaginary,. lol... but what the way?

Answer 45

integral 1/(log^2(x)+1) dx = -1/2 i e^(-i) (e^(2 i) \ Ei(log(x)-i)-Ei(log(x)+i))+constant

Answer 46

that is what wolfram says

Answer 47

and dude, why am I even doing this?
this is hard as hell, I am talking about just basic by part integration, and maybe some partial fractions.

Answer 48

Ei stands for exponential integral

Answer 49

never learned that-:()

Answer 50

\[u=(\ln(x))^2~,~ du = 2\frac{\ln(x)}{x}~,~ dv =dx ~,~ v = x\]

Answer 51

what is up with that picture?? jk

Answer 52

\[\int udv = uv - \int vdu\]\[\int (\ln(x))^2dx = x(\ln(x))^2 -2 \int x\left(\frac{\ln(x)}{x}\right)dx\]

Answer 53

this is how I would go about solving it.

Answer 54

what in the world are you doing?

Answer 55

excuse me

Answer 56

you had the (ln x)^2 + 1 in the denominator

Answer 57

actually, I don't give a .... about this question anyways... tnx for trying Jhanny

Answer 58

\[\int\limits_{}^{} 1 \times (\ln x)^2 dx\]
\[u = (\ln x)^2, u'=2(\ln x)\times \frac{ 1 }{ x },v'=1, v=x\]
\[(\ln x)^2 \times x - \int\limits_{}^{}x \times 2(\ln x) \times \frac{ 1 }{ x } dx\]

Answer 59

Thank you for agreeing @Ankh

Answer 60

DOo you see how it works, @perl?

Answer 61

\[\int (\ln(x))^2dx = x(\ln(x))^2 -2 \int x\left(\frac{\ln(x)}{x}\right)dx\]\[=x(\ln(x))^2 -2\int \ln(x)dx\]

Answer 62

\[\tan ^{-1}(lnx) \] cant be integrated !!

Answer 63

@fbi2015 Can u???

Answer 64

There is no need for trig subs here, @TheLOL

Answer 65

its there above for a question @Jhannybean so I th8 that is what u were talking bout!!

Answer 66

since you don't believe me, check wolfram. http://www.wolframalpha.com/input/?i=int+%28ln%28x%29%29%5E2dx

Answer 67

dont believe u where??

Answer 68

that one is already solved it seems @Jhannybean too bad!!