Anonymous

........3x F(x)=S dt/t=> .........x ..........3x F(x)=[ln(t)]=> ..........x F(x)=ln(3x)-ln(x)=> F(x)=ln(3x/x)=> F(x)=ln(3) is a constant.

Anonymous

Alternately, observe that F'(x) = 0 (and thus F(x) is constant): F'(x) = (1/(3x)) * (3x)' - (1/x) * x' = 3/(3x) - 1/x = 0. I hope this helps!

Anonymous

Integral 1/t dt = ln(t) Integral[x to 3x](1/t)dt = ln(3x)-ln(x) = ln(3)+ln(x)-ln(x) = ln(3) (constant) (whatever x is, it cancels out)

Anonymous

It's C.

Anonymous

.......3x................3x F(x)=∫ 1/t dt=ln|t| |= ........x.................x x∈(0,+oo) are positive hence ..3x lnt |=ln3x-lnx=ln(3)+lnx-lnx=ln(3) Yes! it's a constant ....x

Anonymous

F(x) = ∫(1/t)dt from x to 3x = [ ln(t)] from x to 3x = ln(3x) - ln(x) = ln(3x/x) = ln(3) .......... constant