Water flows into a tank according to the rate F of t equals the quotient of 6 plus t and the quantity 1 plus t , and at the same time empties out at the rate E of t equals the quotient of the natural log of the quantity t plus 2 and the quantity t plus 1 , with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest gallon, is in the tank at time t = 10 minutes.

Question

Steve · Answer

F of t equals the quotient of 6 plus t and the quantity 1 plus t??
How about
F(t)=(6+t)/(1+t)
E(t)=ln(t+2)/(t+1)
Since these are the rates of change, the amount of water will be
V(t)=∫[0,10] F(t)-E(t) dt
=∫[0,10] (6+t)/(1+t)-ln(t+2)/(t+1) dt

Now, (6+t)/(1+t) = 1 + 5/(1+t)
So, ∫(6+t)/(1+t) dt = t + 5ln(1+t)

Now, ln(t+2)/(t+1) does not integrate using elementary functions.
If I parsed your words wrong, then fix things, and you should be able to finish up where I have left off.