NIMCET 2014 Question Paper with solutions

Let l = ∫ $\mathit{x}\frac{{\mathit{e}}^{\mathit{x}}}{\sqrt{1+{\mathit{e}}^{\mathit{x}}}}$ dx
Put 1 + ${\mathit{e}}^{\mathit{x}}$ = ${\mathit{t}}^2$
⇒ ${\mathit{e}}^{\mathit{x}}$ dx = 2 t dt
x = log (${\mathit{t}}^2$ - 1)
∴ l = 2 ∫ log $\frac{\left({\mathit{t}}^2-1\right)}{\mathit{t}}$ t st = 2 ∫ log (${\mathit{t}}^2$ - 1) dt
= 2 $\left[\mathit{t}\mathit{l}\mathit{o}\mathit{g}\left({\mathit{t}}^2-1\right)-2\int \frac{{\mathit{t}}^2}{\mathit{t}-1}\mathit{d}\mathit{t}\right]$
= 2
= 2 $\left[\mathit{t}\mathit{l}\mathit{o}\mathit{g}\left({\mathit{t}}^2-1\right)-2\mathit{t}-\mathit{l}\mathit{o}\mathit{g}\left(\frac{\mathit{t}-1}{\mathit{t}+1}\right)\right]$ + C
= 2x $\sqrt{1+{\mathit{e}}^{\mathit{x}}}$ - 4 $\sqrt{1+{\mathit{e}}^{\mathit{x}}}$ - 2 log $\left(\frac{\sqrt{1+{\mathit{e}}^{\mathit{x}}}-1}{\sqrt{1+{\mathit{e}}^{\mathit{x}}}+1}\right)$ + C
= (2x - 4) $\sqrt{1+{\mathit{e}}^{\mathit{x}}}$ - 2 log $\left(\frac{\sqrt{1+{\mathit{e}}^{\mathit{x}}}-1}{\sqrt{1+{\mathit{e}}^{\mathit{x}}}+1}\right)$ + C
Hence, f(x) = 2x - 4