If ∫2ex+3e−x3ex+4e−xdx=Ax+Blog⁡3e2x+4+C, then

Let∫2ex+3e−x3ex+4e−xdx=∫2e2x+33e2x+4dx 2e2x+3=A3e2x+4+B6e2x⇒ 2e2x+3=(3A+6B)e−2x+4AOn comparing both sides, we get2=3A+6B----i3=4A----iiFrom Eqs. (i) and (ii), woe get 2=94+6B⇒ 6B=2−94=−14  ⇒B=−124 ⇒2e2x+33e2x+4dx=∫33e2x+443e2x+4dx−124∫6e2x3e2x+4dx =34x−124log⁡3e2x+4+C