English

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

Advertisements
Advertisements

Question

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

Sum
Advertisements

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"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Integration - MISCELLANEOUS EXERCISE - 5 [Page 139]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 5 Integration
MISCELLANEOUS EXERCISE - 5 | Q IV. 5) iii) | Page 139

RELATED QUESTIONS

Evaluate : `int x^2/((x^2+2)(2x^2+1))dx` 


Integrate the rational function:

`x/((x^2+1)(x - 1))`


Integrate the rational function:

`x/((x -1)^2 (x+ 2))`


Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`


Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`


Integrate the following w.r.t. x : `2^x/(4^x - 3 * 2^x - 4`


Integrate the following w.r.t. x : `(3x - 2)/((x + 1)^2(x + 3)`


Integrate the following w.r.t. x : `((3sin - 2)*cosx)/(5 - 4sin x - cos^2x)`


Integrate the following w.r.t. x : `(1)/(2sinx + sin2x)`


Integrate the following w.r.t. x : `(5*e^x)/((e^x + 1)(e^(2x) + 9)`


Evaluate:

`int (2x + 1)/(x(x - 1)(x - 4)) dx`.


Evaluate: `int ("x"^2 + "x" - 1)/("x"^2 + "x" - 6)` dx


Evaluate: `int 1/("x"("x"^5 + 1))` dx


Evaluate: `int 1/("x"("x"^"n" + 1))` dx


Evaluate: `int (5"x"^2 + 20"x" + 6)/("x"^3 + 2"x"^2 + "x")` dx


For `int ("x - 1")/("x + 1")^3  "e"^"x" "dx" = "e"^"x"` f(x) + c, f(x) = (x + 1)2.


If f'(x) = `x - 3/x^3`, f(1) = `11/2` find f(x)


`int (3x + 4)/sqrt(2x^2 + 2x + 1)  "d"x`


`int ("d"x)/(x^3 - 1)`


`int xcos^3x  "d"x`


`int (sin2x)/(3sin^4x - 4sin^2x + 1)  "d"x`


Choose the correct alternative:

`int sqrt(1 + x)  "d"x` =


Choose the correct alternative:

`int (x + 2)/(2x^2 + 6x + 5) "d"x = "p"int (4x + 6)/(2x^2 + 6x + 5) "d"x + 1/2 int 1/(2x^2 + 6x + 5)"d"x`, then p = ?


If f'(x) = `1/x + x` and f(1) = `5/2`, then f(x) = log x + `x^2/2` + ______ + c


`int 1/x^3 [log x^x]^2  "d"x` = p(log x)3 + c Then p = ______


Evaluate the following:

`int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`


If f(x) = `int(3x - 1)x(x + 1)(18x^11 + 15x^10 - 10x^9)^(1/6)dx`, where f(0) = 0, is in the form of `((18x^α + 15x^β - 10x^γ)^δ)/θ`, then (3α + 4β + 5γ + 6δ + 7θ) is ______. (Where δ is a rational number in its simplest form)


Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`


Evaluate`int(5x^2-6x+3)/(2x-3)dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×