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

giorgiana1976 | Student

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)

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

We'll combine and eliminate like terms:

2A-2B=1 (5)

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

We'll combine and eliminate like terms:

2A+2B=0 (6)

4A = 1

A = 1/4

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)

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

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)

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

william1941 | Student

If a function d(t) represents the distance. The velocity is given by [f(t+x) - f(t)]/ x. For lim x-->0 we have the instantaneous velocity. This is also equal to the derivative of the function representing distance or d(t)

Here we have d'(t) = f(t) = ln (t) / t (1- ln^4t)

Now we need to find the integral of f(t) and that gives the function of distance.

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

put x = ln t

=> dx = dt/t

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

=> Int [ x / (1- x^4) dx]

=> [ln ( x^2 + 1) - ln (x^2 -1) ]/4

substituting back x = ln t

=> [ln ( (ln t)^2 +1) - ln ( (ln t)^2 -1)] / 4 + C

Therefore the formula for distance is d(t) = [ln ( (ln t)^2 +1) - ln ( (ln t)^2 -1)] / 4 + C