If ∫2ex+3e−x3ex+4e−xdx=Ax+Blog3e2x+4+C, then
A=−34,B=124
A=34,B=−124
A=14,B=124
A=−34,B=14
Let
∫2ex+3e−x3ex+4e−xdx=∫2e2x+33e2x+4dx 2e2x+3=A3e2x+4+B6e2x⇒ 2e2x+3=(3A+6B)e−2x+4A
On comparing both sides, we get
2=3A+6B----i3=4A----ii
From 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−124log3e2x+4+C