# Evaluate: ∫1+logxx(3+logx)(2+3logx) dx - Mathematics

Sum

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"

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Chapter 5: Integration - Miscellaneous Exercise 5 [Page 139]

#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 5 Integration
Miscellaneous Exercise 5 | Q 4.5 | Page 139

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