Question

integrate x tanh^-1 x dx

Answer 1

First thing that comes to mind:\[
\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}} \]So maybe if you solve for \(y\)\[
y = \frac{e^x-e^{-x}}{e^x+e^{-x}}
\]You will get a function that isn't too hard to integrate.

Answer 2

I mean if you solve for \(x\) .

Answer 3

@wio what in my mind is only by parts or tabula maybe?

Answer 4

If you did do it by parts, then you want \(dv\) to be easy to integrate. So you'd have to do something like \(u = tan^{-1}(x) \quad dv = xdx\).

Answer 5

@wio tanh^-1 h there

Answer 6

Yeah, oops.

Answer 7

but, when i tried that way it seem my working does not stop

Answer 8

Can you plug it into the formula then? \[
\int udv = uv-\int vdu
\]

Answer 9

Okay, what did you get so far?

Answer 10

Your work, should definitely stop at some point.

Answer 11

\[
u = \tanh^{-1}(x)\\
du = ? \\
v = ? \\
dv = xdx
\]

Answer 12

\[\frac{ x^2 }{ 2} \tanh^-1 - \frac{ x^2 }{ 2(1-2x) }\]

Answer 13

i left integrate symbol there

Answer 14

@wio \[\int\limits_{}^{}\frac{ x^2 }{ 2-2x62 }\]

Answer 15

it should be x^2 there.,wat would u get?

Answer 16

I'm getting \[
\int \frac{x^2dx}{2-2x^2}
\]as my \(\int vdu\)

Answer 17

What about doing another u substitution now? What about \(u = 1-x^2\)

Answer 18

Or maybe just \(u^2 = 1-x^2\).

Answer 19

there still x above

Answer 20

Then \(x^2 = u^2+1\)?

Answer 21

That would get you: \[
\int \frac{u^2+1}{2u^2}du = \int \frac{u^2}{2u^2} + \frac{1}{2u^2}du
\]

Answer 22

ok

Answer 23

Do you see how to do it now?

Answer 24

@gee can you do the rest?

Answer 25

@wio i think i just want to use the substitution of identity

Answer 26

I learned DETAIL for an easy way to do integration by parts.

Whatever letter is closer to D in detail should be the one you choose as the one multiplied by dx. I'll show you.

D
E - exponentials like e^x
T - trig functions like sin(x)
A - algebraic like x^2
I - inverse trig functions like tan^-1(x)
L- logarithms like ln(x)

So if you see integral of x*e^xdx then you would separate them out as:

x=u and e^xdx=dv

since e^x is closer to D in DETAIL.