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Question
`int ("x" + 1/"x")^3 "dx"` = ______
Options
`1/4 ("x" + 1/"x")^4` + c
`"x"^4/4 + "3x"^2/2 + 3 log "x" - 1/"2x"^2 + "c"`
`"x"^4/4 + "3x"^2/2 + 3 log "x" + 1/"x"^2 + "c"`
`("x" - "x"^-1)^3` + c
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Solution
`int ("x" + 1/"x")^3 "dx"` = `bbunderline("x"^4/4 + "3x"^2/2 + 3 log "x" - 1/"2x"^2 + "c")`
Explanation:
Let I = `int ("x" + 1/"x")^3 "dx"`
`int ("x"^3 + "3x" + 3/"x" + 1/"x"^3)` dx
`= "x"^4/4 + 3 "x"^2/2 + 3 log |"x"| - 1/"2x"^2` + c
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