Given vi and a(v) calculate 1. v(t) 2. time to stop 3. distance to top

[IvyCap]

-->A particle is slowing down from initial velocity v_0=10 with acceleration a=-(1+2v), where v is velocity.
Find
a) v(t)
b) time to stop
c) stopping distance

What I have so far
A) V=Vo+at
v=10+(-1-2v)t
v+2vt=10-t
v(1+2t)=10-t
v(t)=(10-t)/(1+2t)

B) V=0m/s at stop
0=(10-t)/(1+2t)
0=10-t
t=10s

C) This is where I'm stuck, I think I have to integrate V=(10-t)/(1+2t), t=0,10 in order to get my distance covered in t seconds. But I'm not quite sure?

 

[IvyCap]

I think I may have gotten the last part

v=dx/dt

v dt = dx

∫v dt= ∫dx

∫(10-t)/(1+2t) dt = ∫dx t=0 to 10 and x=0 to x

and this is where my integration is so bad I get stuck.