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Question
`int "dx"/(9"x"^2 + 1)= ______. `
Options
`1/3 "tan"^-1(2"x") +"c"`
`1/3 "tan"^-1"x" +"c"`
`1/3 "tan"^-1(3"x") +"c"`
`1/3 "tan"^-1(6"x") +"c"`
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Solution
`1/3 "tan"^-1(3"x") +"c"`
Let I = `int "dx"/(9"x"^2 + 1)`
= `1/9 int "dx"/(("x"^2) +(1/3)^2)`
= `1/9 1/(1/3) "tan"^-1("x"/(1/3)) + "C"`
`= 1/3 "tan"^-1(3"x") + "c"`
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