∫ $\frac{e^x}{\sqrt{5-4e^x-e^{2x}}}$ dx Let $e^x$ = t , $e^x$ dx = dt Now integral I becomes, I = ∫ $\frac{dt}{\sqrt{5-4t-t^2}}$ ⇒ I = ∫ $\frac{dt}{\sqrt{5+4-4-4t-t^2}}$ ⇒ I = ∫ $\frac{dt}{\sqrt{9-(4+4t+t^2)}}$ ⇒ I = ∫ $\frac{dt}{\sqrt{3^2-(t+2)^2}}$ ⇒ I = $\frac{\sin^{-1}(t+2)}{3}$ + C ⇒ I = $\frac{\sin^{-1}(e^x+2)}{3}$ + C OR $\frac{(x-4)e^x}{(x-2)^3}$ dx I = $\int e^x \left( \frac{x-2}{(x-2)^3} - \frac{2}{(x-2)^3} \right)$ dx I = ∫ $e^x \left( \frac{1}{(x-2)^2} - \frac{2}{(x-2)^3} \right)$ dx Thus the given integral is of the form, I = ∫ $e^x$ |f (x) + f' (x)| dx where , f (x) = $\frac{1}{(x-2)^2}$ ; f' (x) = $\frac{-2}{(x-2)^3}$ I = ∫ $\frac{e^x}{(x-2)^2}$ dx - ∫ $\frac{2e^x}{(x-2)^3}$ dx = $\frac{e^x}{(x-2)^2}$ - ∫ $\frac{e^x(-2)}{(x-2)^3}$ dx - ∫ $\frac{2e^x}{(x-2)^3}$ dx + C So, I = $\frac{e^x}{(x-2)^2}$ + C