maths differential eq

dx/dt=((8-5t))/((2/3 t+t^2 ) )
dx= ((8-5t))/((2/3 t+t^2 ) ) dt
∫dx=∫〖((8-5t))/((2/3 t+t^2 ) ) dt〗
x=∫[8/((t^2+2/3 t+1/9)-1/9)-5t/(2/3 t+t^2 )]dt
x=∫[8/((t+1/3)^2-1/9)-5/(2/3+t)]dt
x=∫〖8/((t+1/3)^2-1/9) dt〗-5 ln(2/3+t)
Let 1/3 sinθ=t+ 1/3,get cosθ=√(-9t^2-6t),cosθ dθ=3dt
∫〖8/((t+1/3)^2-1/9) dt〗=∫(8/3 cosθ dθ)/(1/9 (sin^2 θ-1) )=∫(24 cos θ dθ)/(-cos^2 θ )=∫24dθ/〖-cos〗 θ =-24ln 〖(sec θ+tan θ)〗+C=-24 ln 〖((3t+1)/√(-9t^2-6t))〗+C
Therefore,
x=-24 ln ((3t+1)/√(-9t^2-6t))-5 ln (2/3+t)+C=24 ln (√(-9t^2-6t)/(3t+1))-5 ln (2/3+t)+C=ln 〖(√(-9t^2-6t)/(3t+1))^24 〗-ln 〖(2/3+t)^5 〗+C
When x=1,t=0
1= ln 〖(0/1)^24 〗-ln 〖(2/3)^5 〗+C
C= ln 〖(2/3)^5 〗

x = 24 ln (√(-9t^2-6t)/(3t+1))-5 ln (2/3+t)+ln 〖(2/3)^5 〗