Hints on an integration problem...?

Question

integral of: (xln(x)) / sqrt (x^2-1) dx

I would like a couple of hints..like where to start and what to substitute, but you don't have to solve it all for me. If I need further help, I'll let you know.

hmmm math_kp, i got confused =[
where does this come from? :
udv/dx = d(uv)/dx - vdu/dx

and this part...i am confused..do i add another variable, t?
x/sqrt(x^2-1) is integrable by putting x^2-1 = t

Answer 1

my approcah would be taking differentiation of product formula

as d(uv)/dx = udv/dx + vdu/dx
we get

udv/dx = d(uv)/dx - vdu/dx

take u = ln x
and v = x/sqrt(x^2-1)
ln x is not integrable easily so take u = ln x so du/dx = 1/x
and v = x/sqrt(x^2-1) as this is integrable by substitution

x/sqrt(x^2-1) is integrable by putting x^2-1 = t
the xdx = dt

so x/sqrt(x^2-1) = dt/2t^(1/2) = t^(1/2)= sqrt(x^2-1)

you proceed along the line and you should be able to solve.
In case you cannot then please let me know

Answer 2

Try using integration by parts:
u = ln x dv = x dx/x^2 -1.
Unfortunately, the integrals you get are not
elementary and neither is the original integral.
It's solution involves the dilogarithm function. Why? Well,
carrying out the idea above, we get
du = 1/x dx v = 1/2* ln(x^2-1)
So, if we call your original integral I,
I = 1/2*[ ln(x) ln(x^2 -1) - int ln(x^2-1)/x dx]
So let's tackle J = int ln(x^2-1).x dx.
At first this seems hard to deal with. But, look,
write ln(x^2 -1) = ln( (x+1)(x-1) ) = ln(x+1) + ln(x-1).
So J = int ln (x+1)/x + int ln (x-1)/x.
Now I used integrals.wolfram.com to do these last two.
It gave me
int ln (x+1)/x dx = -dilog(-x)
and
int ln(x-1)/x dx = ln(x)( ln(x-1) - ln(1-x) ) -dilog(x).
It can be proved that these last 2 integrals
really do involve the dilog function, i.e.,
they are really nonelementary.
Finally, the other ideas previously mentioned
also lead to nonelementary integrals.

Answer 3

Put u = ln(x) and dv = x/sqrt(x^2 -1)) du. By parts, your integral is u*v - Int v du. Since d/dx(x^2 -1) = 2x, we get v = sqrt(x^ 2 -1). And vdu = sqrt(x^ 2 -1)*(1/x) dx. To integrate vdu, put x = cosh(t). Then, vdu = senh(t)/cosh(t). senh(t) dt = dt/cosh(t). Now it`s eeasir, right?

Answer 4

put (x^2-1)^(1/2) = t
square both side
x^2 -1 =t^2 0r x =sq rt(1+t^2) or ln(x)=e^{sq rt(1+t^2)}
x^2=1+t2 differentiate both sides
2x dx =2t.dt
x dx = t dt....................ii
on simplifying get the integration of e^{sq rt(1+t^2)} which is very simple and put back sq rt(1+t^2)=x and t=(x^2+1)^(1/2)
Second method put x=sec t
Therfore dx =sec t tan t dt and ln(x) = e^sec t
and x^2-1=sec^2 t -1=tan^2(t)
sqrt(sec^2(t)-1)=tan t
Rest do your self