Integrate by substitution dx/4x+3?

Answer 1

You do this in the following way:

substitute u=4x+3

the differentiate to get

du/dx=4

so that dx=du/4

then integrate

du/4u

which should give

(ln u) /4

then you can reverse the substitution to give

(ln 4x+3)/4

Answer 2

let p = 4x+3
dp/dx = 4 => dx = dp/4

then INT[dx/(4x+3)] = INT[ 1/p4 dp] = 1/4ln[p] + c= 1/4ln[4x+3] + c

Answer 3

dx/4x+3=-1x