Question

Evaluate the following definite integrals: ${\displaystyle {\int }_1^2\frac{dx}{x^2+6x+5}}$

Answer 1

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

= ${\displaystyle {\int }_1^2\frac{dx}{x^2+6x+9-9+5}}$

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

= ${\displaystyle \frac{1}{2x2}{\left[\log \left|\frac{x+3-2}{x+3+2}\right|\right]}_1^2}$

= ${\displaystyle \frac{1}{4}{\left[\log \left|\frac{x+1}{x+5}\right|\right]}_1^2}$

= ${\displaystyle \frac{1}{4}[\log \frac{3}{7}-\log \frac{2}{6}]}$

= ${\displaystyle \frac{1}{4}\log (\frac{3}{7}\times \frac{6}{2})}$

∴ I ${\displaystyle \frac{1}{4}\log \left(\frac{9}{7}\right)}$.