Integrate 2x/(x^2+3) using the substitution u=x^2+3
u=x2 + 3
du/dx=2x
dx=du/2x
2x/(x2+3) dx becomes (2x/u) * (du/2x)
the 2x terms cancel out giving 1/u du
this integrates to ln(u)+c becoming ln(x2+3)+c
TS
u=x2 + 3
du/dx=2x
dx=du/2x
2x/(x2+3) dx becomes (2x/u) * (du/2x)
the 2x terms cancel out giving 1/u du
this integrates to ln(u)+c becoming ln(x2+3)+c
