Question

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

Answer 1

${\int }_0^1\frac{x}{x^2+1}dx$
Let $x^2+1=t\Rightarrow 2xdx=dt$
When $x=0,t=1$ and when $x=1,t=2$
$\therefore {\int }_0^1\frac{x}{x^2+1}dx=\frac{1}{2}{\int }_1^2\frac{dt}{t}$
$=\frac{1}{2}[\log |t|{]}_1^2$
$=\frac{1}{2}[\log 2-\log 1]=\frac{1}{2}\log 2$