Question

Evaluate the following integral
${\int }_2^3\frac{1}{x^2+6x-7}dx$

Answer 1

Formula: ${\int }_a^b\frac{1}{x+a}dx=\ln (x+a)$

${\int }_2^3\frac{1}{x^2+6x-7}dx$

$\Rightarrow {\int }_2^3\frac{1}{x^2+6x+9-9-7}dx$

$\Rightarrow {\int }_2^3\frac{1}{(x+3{)}^2-16}dx$

$\Rightarrow {\int }_2^3\frac{1}{(x+3{)}^2-4^2}$

$\Rightarrow {\int }_2^3\frac{1}{(x+3-4)(x+3+4)}$

$\Rightarrow {\int }_2^3\frac{1}{(x-1)(x+7)}dx$

$\Rightarrow {\int }_2^3\frac{1}{8}[\frac{1}{x-1}-\frac{1}{x+7}]dx$

$\Rightarrow \frac{1}{8}|\ln (x-1)-\ln (x+7){|}_2^3$

$\Rightarrow \frac{1}{8}[(\ln 2-\ln 1)-(\ln 10-\ln 9)]$

$\Rightarrow \frac{1}{8}[\ln 2-0-\ln 10+\ln 9]$

$\Rightarrow \frac{1}{8}[\ln \frac{2\times 8}{10}]$

$\Rightarrow \frac{1}{8}\ln 1.8$

Therefore, ${\int }_2^3\frac{1}{x^2+6x-7}dx=\frac{1}{8}\ln 1.8$