Question

Evaluate the integral
${\int }_4^5\frac{x+1}{(x^2-9)}dx$

Answer 1

Given : ${\int }_4^5\frac{(x+1)}{(x^2-9)}dx$

$I={\int }_4^5\frac{(x+1)}{(x^2-9)}dx={\int }_4^5\frac{x}{(x^2-9)}dx+{\int }_4^5\frac{dx}{(x^2-9)}I=\frac{1}{2}{\int }_4^5\frac{2x}{(x^2-9)}dx+{\int }_4^5\frac{dx}{(x^2-9)}I=\frac{1}{2}\log {\left|\right(x^2-9\left)\right|}_4^5+\frac{1}{6}\log {\left|\frac{x-3}{x+3}\right|}_4^{5}I=\frac{1}{2}[2\log 4-\log 7]+\frac{1}{6}(\log \frac{2}{8}-\log \frac{1}{7})I=\log 4-\frac{1}{6}\log 4-\frac{1}{2}\log 7+\frac{1}{6}\log 7I=\frac{5}{6}\log 4-\frac{1}{3}\log 7$

Hence the correct answer is $\frac{5}{6}\log 4-\frac{1}{3}\log 7$