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प्रश्न
`int"x"^"n"."log" "x" "dx"`
बेरीज
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उत्तर
`int"x"^"n"."log" "x" "dx"`
= `int"log" "x" "x"^"n" "dx"`
`int"u"."v" "dx" = "u" int "v" "dx" - int ("du"/"dx") [int "v dx"] "dx"`
= `"log x" int "x"^"n" "dx" - int ["d"/"dx" ("log x")int "x"^"n" "dx"] "dx"`
= `"log x" xx ("x"^("n" + 1))/("n" + 1) - int 1/"x".("x"^("n" + 1))/("n" + 1) "dx"`
= `("x"^("n" + 1) "log x")/("n" + 1) - 1/("n" + 1) int"x"^"n" "dx"`
= `("x"^("n" + 1) "log x")/("n" + 1) - ("x"^("n" + 1))/("n" + 1)^2 + "C"`
= `("x"^("n" + 1) "log x")/("n" + 1) ["log x" - 1/("n" + 1)] + "C"`
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