Question

Evaluate the integral
${\int }_1^2\frac{1}{(x^2+x)}dx$

Answer 1

Given :${\int }_1^2\frac{1}{(x^2+x)}dx$

$I={\int }_1^2\frac{dx}{(x^2+x)}={\int }_1^2\frac{dx}{x^2(1+\frac{1}{x})}$

Let :

$z=1+\frac{1}{x}x\to 1t\to 2-dt=\frac{dx}{x^2}x\to 2t\to \frac{3}{2}I={\int }_2^{\frac{3}{2}}-\frac{dt}{t}=-{\left[\log \left|t\right|\right]}_2^{\frac{3}{2}}=-\log \left|\frac{3}{2}\right|+\log 2=\log \left(\frac{4}{3}\right)$

Hence the correct answer is $\log \left(\frac{4}{3}\right)$