Evaluate $\int\limits_{0}^{1} \frac{x}{\sqrt{1+x^2}} dx$

$\sqrt{2}-1$

The correct answer is Option (3) → $\sqrt{2}-1$

Let $I = \int\limits_{0}^{1} \frac{x}{\sqrt{1+x^2}} dx$

Put $1 + x^2 = t^2$

$\Rightarrow 2x \, dx = 2t \, dt$

$\Rightarrow x \, dx = t \, dt$

$∴I = \int\limits_{1}^{\sqrt{2}} \frac{t \, dt}{t}  [\text{when } x = 0, \text{ then } 1 + (0)^2 = t^2, t = 1]$

$[\text{when } x = 1, \text{ then } 1 + 1^2 = t^2, t = \sqrt{2}]$

$= [t]_1^{\sqrt{2}} = \sqrt{2} – 1$