**Evaluate: **`int (1 + log "x")/("x"(3 + log "x")(2 + 3 log "x"))` dx

#### 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"`