Q: Find the arc length of the curve y=ln(x) where x ranges from √3 to √15.
I think I am stuck in calculation part.
The answer is 2+ln(3)−12ln(5). But I can't derive that from my last line.
help me, please.
Answer
Your problem now is how to evaluate the integral ∫√15√3√x2+1xdx.
Let
F(x)=∫√x2+1xdx.
Let x=tanθ. All your computations after this substitution are all correct.
Note that cscθ=√x2+1x cotθ=1x and
secθ=√x2+1.
Hence, we get
F(x)=−ln(√x2+1+1x)+√x2+1+C.
Thus,
∫√15√3√x2+1xdx=F(√15)−F(√3)=[−ln(5√15)+4+C]−[−ln(3√3)+2+C]=ln(3√3)−ln(5√15)+2=ln[3√3÷5√15]+2=ln[3√3⋅√155]+2=ln[31⋅√55]+2=ln[31⋅1√5]+2=ln(3√5)+2=ln3−ln√5+2=ln3−12(ln5)+2.
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