# Evaluate the following: $\int\limits_0^1\frac{xdx}{\sqrt{1+x^2}}$

Toolbox:
• $\int \limits_a^b x^n=\bigg[ \large \frac{x^{n+1}}{n+1}\bigg]_a^b$
Let $I=\large\int\limits_0^1\frac{xdx}{\sqrt{1+x^2}}$
Put $1+x^2=t$
On differentiating with respect to t we get,
$2xdx=dt =>xdx=dt/2$
When we substitute for t, the limits also change
When $x=0, \qquad 1+0=t=>t=1$
When $x=1, \qquad 1+1=t=>t=2$
On substituting this in I we get,
Therefore $I=\large\frac{1}{2} \int \limits^2_1 \frac{dt}{\sqrt t}$