Question

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

Answer 1

${\int }_0^1\frac{x}{x^2+1}.dx$

Put $t=x^2+1$

Differentiating $w.r.t.x$

$\frac{dt}{dx}=2x\Rightarrow \frac{dt}{2}=xdx$

The new limits are

When $x=0\Rightarrow t=0^2+1=1$

When $x=1\Rightarrow t=1^2+1=2$

Therefore,

${\int }_0^1\frac{x}{x^2+1}.dx$

$=\frac{1}{2}{\int }_1^2\frac{dt}{t}$

$=\frac{1}{2}[\log |t|{]}_1^2$

$=\frac{1}{2}[\log |2|-\log |1|]$

$=\frac{1}{2}[\log |2|-0]$

$=\frac{1}{2}\log 2$