Question

Evaluate the definite integrals.
${\int }_2^3\frac{x}{x^2+1}dx.$

Answer 1

Let $I={\int }_2^3\frac{x}{x^2+1}dx$
Put $x^2+1=t\Rightarrow 2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2x}$
$\therefore I={\int }_2^3\frac{x}{t}\frac{dt}{2x}=\frac{1}{2}{\int }_2^3\frac{1}{t}dx=\frac{1}{2}[log|t|{]}_2^3=\frac{1}{2}[log|x^2+1|{]}_2^3(\because t=x^2+1)=\frac{1}{2}[log|3^2+1|-log|2^2+1|]=\frac{1}{2}[log|10|-log|5|]=\frac{1}{2}log\left|\frac{10}{5}\right|=\frac{1}{2}log2(\because loga-logb=log\frac{a}{b})$