Question

an object moves in a straight line with a velocity function: v(t)= (x+3)/((x^2)+3x+2) ft/sec. Find the distance traveled from rest to 4 seconds later.

Answer 1

First thing here is knowing that the distance travelled, is the area under a velocity time graph. So the total distance travelled is equal to the integral of your function from 0 to 4, as so: \[\int\limits_{t=0}^{4}\frac{ x+3 }{ x^2 +3x +2 }dx\] Now notice the bottom of it factors into (x+1)(x+2). So now we can express this as partial fractions. We get:

\[\frac{ x+3 }{ (x+1)(x+2) }= \frac{ A }{ x+1 }+ \frac{ B }{ x+2 }\]\[x+3= A(x+2) + B(x+1)\] Now first let x=-1 to get rid of the B, then let x=-2 to get rid of A and we see that overall A=2 and B=-1. Overall: \[\int\limits_{x=0}^{4}\frac{ x+3 }{ x^2+3x+2}dx = \int\limits_{x=0}^{4}\frac{ 2 }{ x+1 } - \frac{ 1 }{ x+2 }dx\]So the integral is \[[2\ln(x+1) -\ln(x+2)]_{0}^{4}\] Evaluating this gives us. \[[2\ln(5) - \ln(6)] - [2\ln(1) - \ln(2)]\] ln(1) being 0, then ln(6)=ln(3) + ln(2) thus the ln(2)'s cancel. \[\ln(25) -\ln(3) - \ln(2) +\ln(2)\] Overall answer is \[\ln(\frac{ 25 }{ 3 })\]Hope that this helped :)