Question

Evaluate: $\underset{1}{\overset{2}{\int }}\frac{dx}{x^2+6x+5}$

Answer 1

$\underset{1}{\overset{2}{\int }}\frac{dx}{x^2+6x+5}$

$\underset{1}{\overset{2}{\int }}\frac{dx}{x^2+5x+x+5}=\underset{1}{\overset{2}{\int }}\frac{dx}{(x+5)(x+1)}$

$=\frac{1}{4}\underset{1}{\overset{2}{\int }}\frac{(x+5)-(x+1)}{(x+5)(x+1)}dx$

$=\frac{1}{4}\underset{1}{\overset{2}{\int }}\frac{dx}{x+1}-\frac{1}{4}\underset{1}{\overset{2}{\int }}\frac{dx}{x+5}$

$=\frac{1}{4}\log (x+1)\underset{1}{\overset{2}{\int }}-\frac{1}{4}\log (x+5)\underset{1}{\overset{2}{\int }}$

$=\frac{1}{4}[\log 3-\log 2]-\frac{1}{4}\left[\log \right(7)-\log (6\left)\right]$

$=\frac{1}{4}\log \left(3/2\right)-\frac{1}{4}\log \left(7/6\right)$

$=\frac{1}{4}\left[\log (\frac{3}{2}\times \frac{6}{7})\right]$

$=\frac{1}{4}\log \frac{9}{7}$