help?

$\LARGE \int \frac{\ln x -4}{x(\ln ^2 x + 3)} dx$ show steps pleeeeaaasssee...i think it has something to do with the inverse trig functions

Question

help?

$\LARGE \int \frac{\ln x -4}{x(\ln ^2 x + 3)} dx$ show steps pleeeeaaasssee...i think it has something to do with the inverse trig functions

lgbasallote · Answer

@Callisto ? you're the only one online i know that knows calculus lol

lgbasallote · Answer

@Mimi_x3 !!! help :P

mimi_x3 · Answer

Try let u=logx

mimi_x3 · Answer

u=lnx in this case.

lgbasallote · Answer

$\LARGE \int \frac{u-4}{u^2 + 3} du$

mimi_x3 · Answer

break it up into two fractions.

lgbasallote · Answer

$\LARGE \int \frac{u}{u^2+3}du - \int \frac{4}{u^2 + 3}du$

mimi_x3 · Answer

Yes! 
The first one is a substitution; and the next is a trig inverse.

mimi_x3 · Answer

Are you able to do it?

lgbasallote · Answer

let me see...

mimi_x3 · Answer

well, they are both easy; if need help just ask me. xD

lgbasallote · Answer

easy for you :P i still need practice to see these things easier

mimi_x3 · Answer

okay sorry..well; just try it, you can do it!

lgbasallote · Answer

$\LARGE \ln (\sqrt{u^2 +3}) - \frac{4\sqrt 3 \tan^{-1} u}{3\sqrt 3}$

lgbasallote · Answer

lol ignore the $\sqrt 3$'s  forgot to cancel them

lgbasallote · Answer

am i right @Mimi_x3 ?

mimi_x3 · Answer

wait..something looks wrong..

lgbasallote · Answer

:O

mimi_x3 · Answer

Congratulations! Correct! ^_^

mimi_x3 · Answer

don't forget to sub back the x!

lgbasallote · Answer

yayyy!!! but how come when i input the answer in wolfram and took the derivative the second term says 4/3(u^2 +1) is that the same?

mimi_x3 · Answer

why did you take the derivative...?

lgbasallote · Answer

to see if it will come out as the integral

lgbasallote · Answer

because taking the integral of that from the start results in crazy manipulations

mimi_x3 · Answer

i don't think that it will work since it's a trig inverse..

mimi_x3 · Answer

wait..which integral are you talking about $$\int\limits\frac{4}{x^{2}+3}  dx$$??

lgbasallote · Answer

oh wait lol it should be $\LARGE \frac{4 \tan^-1 (\frac {u}{sqrt 3})}{sqrt 3}$

lgbasallote · Answer

wonder where i went wrong

mimi_x3 · Answer

you did not do it wrong..

mimi_x3 · Answer

your answer was like that..

lgbasallote · Answer

no...my sqrt 3 was not in the argument...

lgbasallote · Answer

hah lol found it...forgot to put over sqrt 3 in the argument lol

lgbasallote · Answer

it was a simple typo =))

mimi_x3 · Answer

lol, i was going to type the solution up.. xD 

So need anymore help?

lgbasallote · Answer

yeah :DD just some tips though....

mimi_x3 · Answer

Okay; what tips..?

lgbasallote · Answer

i meant hints while solving...ill post another problem

mimi_x3 · Answer

alrighty! try me; too tired now..