What is the function of distance if the function of velocity is f(t)=ln(t)/t(1-ln^4t)

The velocity is the derivative of distance, with respect
to time:

v = ds/dt

vdt =
ds

We'll integrate both
sides:

Int vdt = Int ds

Int
ln(t)dt/t(1-ln^4t) = s

We notice that if we'll re-write the
function, we'll have:

Int
[ln(t)/(1-ln^4t)]*(dt/t)

If we'll substitute ln t = u and
we'll differentiate, we'll get:

dt/t =
du

We'll re-write the integral, changing the
variable:

Int u
du/(1-u^4)

We'll write the ratio u/(1-u^4) as an algebraic
sum of elemntary fractions:

u/(1-u^4) =
u/(1-u)(1+u)(1+u^2)

u/(1-u)(1+u)(1+u^2) = A/(1-u) + B/(1+u)
+ (Cu+D)/(1+u^2)

We'll multiply the ratios from the right
side, so that w'ell obtain LCD.

u = A(1+u)(1+u^2) +
B(1+u^2)(1-u) + (Cu+D)(1-u^2)

u = A + Au^2 + Au + Au^3 + B
- Bu + Bu^2 - Bu^3 + Cu - Cu^3 + D - Du^2

The
correspondent coefficients from both sides have to be
equal:

 u = u^3(A-B-C) + u^2(A+B-D) + u(A-B+C) +
A+B+D

A-B-C=0 (1)

A+B-D=0
(2)

A-B+C=1 (3)

A+B+D=0
(4)

We'll add (1) +
(3):

A-B-C+A-B+C=1

We'll
combine and eliminate like terms:

2A-2B=1
(5)

We'll add (2) +
(4):

A+B-D+A+B+D=0

We'll
combine and eliminate like terms:

2A+2B=0
(6)

We'll add (5)+(6):

4A =
1

A =
1/4

We'll add (1) +
(2):

A-B-C+A+B-D=0

We'll
substitute A and we'll eliminate like terms:

1/2 - C - D =
0

C+D = 1/2 (7)

We'll add (3)
+ (4):

A-B+C+A+B+D=1

We'll
substitute A and we'll eliminate like terms:

1/2 + C+D =
1

C+D = 1 - 1/2

C+D =
1/2

We'll add (2) +
(3):

A+B-D+A-B+C=1

We'll
substitute A and we'll eliminate like terms:

1/2 + C - D =
1

C - D = 1 - 1/2

C - D = 1/2
(8)

We'll add (7) + (8):

2C =
1

C =
1/2

D =
0

B =
-1/4

The integrand will
become:

u/(1-u^4) = 1/4(1-u) - 1/4(1+u) +
u/2(1+u^2)

Int  udt/(1-u^4)=(1/4)Int du/(1-u) - (1/4)Int
du/(1+u)+(1/2)Int udu/(1+u^2)

Int 
udt/(1-u^4)=(1/4)ln|(1-u)/(+u)| +
(1/4)ln(1+u^2)+C

The expression of distance
is:

s =
(1/4)ln|(1-lnt)/(1+lnt)| + (1/4)ln[1+(ln t)^2]+C