Question

what the integration of 1/(1-x^2)*(4+tanh"inverse"x)^(1/2)

Answer 1

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{1-x^2}(4+\tanh^{-1}x)~dx}\) like this?

Answer 2

use u substitution, \(\large\color{slate}{\displaystyle u=4+\tanh^{-1}x}\)

Answer 3

i bet it is in the denominator

Answer 4

no its integaration 1/((1-x^2)*(4+tanh"inverse"x)^(1/2))

Answer 5

that way you can make \(u=arctanh(x)\) and do a u-sub in one step
i could be wrong

Answer 6

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1-x^2)(4+\tanh^{-1}x)}~dx}\)

still do the same u substitution

Answer 7

yes
give me the full answer to make sure that i correctly solve

Answer 8

we don't give answers, or at least try not to.
We can help you out though...

Answer 9

\[\int\frac{dx}{(1-x^2)\sqrt{(4-tanh^{-1}(x)}}\]

Answer 10

just a guess, maybe it is something else

Answer 11

Your problem is?
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1-x^2)(4+\tanh^{-1}x)}~dx}\)

But no matter where \(\large\color{slate}{\displaystyle 4+\tanh^{-1}x}\)
in the root, just in denominator, or multiplied times the fraction,

Do: \(\large\color{blue}{\displaystyle u=4+\tanh^{-1}x}\)

Answer 12

thank you all for helping

Answer 13

(I am still not clear about what is exactly the problem )

Answer 14

is it the last thing I suggested?

Answer 15

one more guess
\[\huge \int\frac{dx}{(1-x^2)\sqrt{(4+tanh^{-1}(x)}}\]

Answer 16

i solved the problem thanks all

Answer 17

what was the problem?
what was the answer you got?