Question

Evaluate the integrals using substitution.
${\int }_0^1\frac{x}{x^2+1}dx.$

Answer 1

Let $I={\int }_0^1\frac{x}{x^2+1}dx$
Put $x^2+1=t\Rightarrow 2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2x}$
For limit, when x=0 $\Rightarrow$ t =1 and when $x=1\Rightarrow t=2(\because t=x^2+1)$
$\therefore I={\int }_1^2\frac{x}{t}\frac{dt}{2x}=\frac{1}{2}{\int }_1^2\frac{1}{t}dt=\frac{1}{2}[log|t|{]}_1^2=\frac{1}{2}[log|2|-log|1|]=\frac{1}{2}log2(\because log1=0)$