Question

सोडवा: `int_1^2 (dx)/(x^2 + 6x + 5)`

Answer 1

समजा I = `int_1^2 1/(x^2 + 6x + 5)`

= `int_1^2 (dx)/(x^2 + 6x + 9 - 9 + 5)`

= `int_1^2 (dx)/((x + 3)^2 - 4)`

= `int_1^2  (dx)/((x + 3)^2 - (2)^2)`

= `1/(2 xx 2)[log|(x + 3 - 2)/(x + 3 + 2)|]_1^2`

= `1/4[log|(x + 1)/(x + 5)|]_1^2`

= `1/4[log(3/7) - log(2/6)]`

= `1/4 log(3/7 xx 6/2)`

∴ I = `1/4 log(9/7)`