Question

Solve:

${\int }_1^2\frac{dx}{x^2+5x+6}=$

Answer 1

${\int }_1^2\frac{dx}{x^2+5x+6}$

$={\int }_1^2\frac{dx}{x^2+2x+3x+6}={\int }_1^2\frac{dx}{x(x+2)+3(x+2)}$

$={\int }_1^2\frac{dx}{(x+3)(x+2)}$

$={\int }_1^2\frac{(x+3)-(x+2)}{(x+3)(x+2)}dx$ $[\because (x+3)-(x+2)=1]$

$={\int }_1^2\frac{x+3}{(x+3)(x+2)}-\frac{(x+2)}{(x+3)(x+2)}dx$

$={\int }_1^2\frac{1}{x+2}-\frac{1}{x+3}dx$

$=[log|x+2|-log|x+3|{]}_1^2$

$=log\frac{|x+2|}{|x+3|}{]}_1^2$ $[\because loga-logb=log\frac{a}{b}]$

$=log\left|\frac{2+2}{2+3}\right|-log\left|\frac{1+2}{1+3}\right|$

$=log\left|\frac{4}{5}\right|-log\left|\frac{3}{4}\right|=log|\frac{4}{5}\times \frac{4}{3}|=log\left|\frac{16}{15}\right|$