∫(x+1)/(x2+2)2x−3​dx need help with this question. can you show the process​

do you need to include the wiggle infront of the first bracket? if not;

(x+1) ÷ [(x^2+2) x (2x-3dx)]

x^2 x 2x = 2x^3

x^2 x -3dx = -3dx^3

2 x 2x = 4x

2 x -3dx = - 6dx

i cant find a way to make it equal 0 so i think the answer is just

x+1 over 2x^3 - 3dx^3 + 4x - 6dx as a fraction

a = 11 : 4

b = 38

c = 1 for every 6

step-by-step explanation: