Question
Evaluate the following:
$\int\limits^1_0\frac{\text{x}}{\sqrt{1+\text{x}^2}}\text{dx}$

Answer

Let $\text{I}=\int\limits^1_0\frac{\text{x}}{\sqrt{1+\text{x}^2}}\text{dx}$
Put $1+\text{x}^2=\text{t}^2$
$\Rightarrow\ 2\text{xdx}=2\text{tdt}$
$\Rightarrow\ \text{xdx}=\text{tdt}$
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi,$ then $\text{t}\rightarrow\sqrt{2}$
Substituting $1+\text{x}^2=\text{t}^2$ and $\text{xdx}=\text{tdt}$ in I, we get
$\therefore\ \text{I}=\int\limits^{\sqrt{2}}_0\frac{\text{tdt}}{\text{t}}$
$=[\text{t}]^{\sqrt{2}}_1=\sqrt{2}-1$

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