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