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प्रश्न
Evaluate : ∫ log (1 + x2) dx
उत्तर
Let I = ∫ log (1 + x2) dx
Put x = tan θ
∴ dx = sec2 θ d θ
`therefore "I" = int "log" (1 + "tan"^2 theta) . "sec"^2 theta "d" theta`
`= int "log"("sec"^2 theta) . "sec"^2 theta "d" theta`
`= 2 int "log" (sec theta) . "sec"^2 theta "d" theta`
`= 2 ["log" ("sec") theta . int "sec"^2 theta "d" theta - int ["d"/("d" theta) "log" ("sec" theta) . int "sec"^2 theta "d" theta] "d" theta]`
`= 2 xx ["log" ("sec" theta) . "tan" theta - int ("sec" theta . "tan" theta)/("sec" theta) . "tan" theta "d" theta]`
`= 2 ["log" ("sec" theta) . "tan" theta - int "tan"^2 "d" theta]`
`= 2 ["tan" theta . "log" ("sec" theta) - int ("sec"^2 theta - 1) "d" theta]`
`= "tan" theta . "log" ("sec"^2 theta) - 2 "tan" theta + 2 theta + "C"`
`= "x" . "log" (1 + "x"^2) - "2x" + 2 "tan"^-1 "x" + "C"`
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