Question

Evaluate
${\int }_1^3(4x+\frac{1}{x}+1)dx$

Answer 1

Consider, $I={\int }_1^3(4x+\frac{1}{x}+1)dx$

$={[\frac{4x^2}{2}+\ln x+x]}_1^3$

$={[2x^2+lnx+x]}_1^3$

$=2(3{)}^2+ln3+3-(2(1{)}^2+ln1+1)$

$=21+ln3-\left(3\right)=18+ln3$

$=18+ln3$