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Question
Find the integrals of the following:
`1/(6x - 7 - x^2)`
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Solution
`int 1/(6x - 7 - x^2) "d"x = int ("d"x)/(- 7 - x^2 + 6x)`
= `int ("d"x)/(- 7 - (x^2 - 6x))`
= `int ("d"x)/(- 7 - [(x - 3)^2 - 3^2]`
= `int ("d"x)/(- 7 - (x - 3)^2 + 9)`
= `int ("d"x)/(2 - (x - 3)^2`
= `int ("d"x)/((sqrt(2))^2 - (x - 3)^2)`
Put x – 3 = t, dx = dt
= `int "dt"/((sqrt(2)^2 - "t"^2)`
= `1/(2 xx (sqrt(2))) log |(sqrt(2) + "t")/(sqrt(2) - "t")| + "c"`
= `1/(2 xx sqrt(2)) log |(sqrt(2) + (x - 3))/(sqrt(2) - (x - 3))| + "c"`
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