Question

please help

Answer 1

\[\int\limits \log(logx+2)dx\]

Answer 2

@rational

Answer 3

i'm thinking integration by parts.\[\int \ln (\ln x + 2) dx = x\ln(\ln x + 2) - \int x\cdot \frac{1}{\ln x+2}\cdot\frac{1}{x} dx\]

Answer 4

i think we have to do by substitution
im not getting what to subsitute

Answer 5

LIATE or LIPETs for usub.

Answer 6

u = ln(lnx + 2) and dv = dx

Answer 7

hey jhaanybean whats P?

Answer 8

LIATE

Log > inverse > algebraic (polynomials) > Trig > Exponential

LIPETs

Log > Inverse > Polynomials (algebraic) > Exponential > Trig

Answer 9

Algebraic expressions are not necessarily polynomials.

Answer 10

Pick your poison.

Answer 11

Well true... but in this little acronym that's what theyre categorized as. that's why I stick with LIATE.

Answer 12

i knew only about LIATE

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @ParthKohli
i'm thinking integration by parts.\[\int \ln (\ln x + 2) dx = x\ln(\ln x + 2) - \int x\cdot \frac{1}{\ln x+2}\cdot\color{red}{\frac{1}{x}} dx\]
\(\color{blue}{\text{End of Quote}}\)

Haha I just had a brain fart and forgot about this little bugger.

Answer 14

@rvc Where did you get this problem?

Answer 15

The second integral is not expressible in elementary terms.

Answer 16

Yeah, I checked WolframAlpha and it's the Ei function.

Answer 17

either \(\int\limits \log(\log(x+2))dx \text{ or } \int\limits \log(\log(x)+2)dx\) you're in trouble haha. You can still integrate this though with a power series.

Answer 18

wait
sorry i made a mistake in posting
its logxlog(x+2)

Answer 19

Oh my...

Answer 20

Let's start over.

Answer 21

im extermely sorry guys
yes lets start

Answer 22

hmm
so what to substitute

Answer 23

OK, nope. I tried it all and this integral is also pretty hard.

Answer 24

@rational here help

Answer 25

Are you sure that this is the question?