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Question
Evaluate: `int (1 + log "x")/("x"(3 + log "x")(2 + 3 log "x"))` dx
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Solution
Let I = `int (1 + log "x")/("x"(3 + log "x")(2 + 3 log "x"))` dx
Put log x = t
∴ `1/"x"` dx = dt
∴ I = `int (1 + "t")/((3 + "t")(2 + "3t"))` dt
Let `(1 + "t")/((3 + "t")(2 + "3t")) = "A"/("3 + t") + "B"/(2 + "3t")`
∴ 1 + t = A(2 + 3t) + B(3 + t) ...(i)
Putting t = – 3 in (i), we get
1 -3 = A(2 - 9) + B(0)
∴ - 2 = A (- 7)
∴ A = `2/7`
Putting t = `- 2/3` in (i), we get
`1 - 2/3 = "A"(0) + "B"(3 - 2/3)`
∴ `1/3 = "B"(7/3)`
∴ B = `1/7`
∴ `("1+t")/(("3 + t")("2 + 3t")) = (2/7)/("3 + t") + (1/7)/(2 + "3t")`
∴ I = `int ((2/7)/("3 + t") + (1/7)/("2 + 3t"))` dt
`= 2/7 int 1/(3+"t") "dt" + 1/7 int 1/(2 + "3t")` dt
`= 2/7 log |3 + "t"| + 1/7 * (log |2 + "3t"|)/3` + c
∴ I = `2/7 log |3 + log "x"| + 1/21 log |2 + 3 log "x"| + "c"`
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