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प्रश्न
Evaluate: `int 1/(2"x" + 3"x" log"x")` dx
उत्तर
Let I = `int 1/(2"x" + 3"x" * log"x")` dx
`= int 1/("x"(2 + 3 log "x"))` dx
Put 2 + 3 log x = t
∴ `3 * 1/"x" "dx"` = dt
∴ `1/"x" "dx" = 1/3 "dt"`
∴ I = `1/3 int 1/"t" * "dt"`
`= 1/3` log |t| + c
∴ I = `1/3` log |2 + 3 log x| + c
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