हिंदी

Integrate the following w.r.t.x: x+5x3+3x2-x-3

Advertisements
Advertisements

प्रश्न

Integrate the following w.r.t.x: `(x + 5)/(x^3 + 3x^2 - x - 3)`

योग
Advertisements

उत्तर

Let I = `int (x + 5)/(x^3 + 3x^2 - x - 3)*dx`

= `int (x + 5)/(x^2(x + 3) - (x + 3))*dx`

= `int (x + 5)/((x + 3)(x^2 - 1)`

= `int (x + 5)/((x + 3)(x - 1)(x + 1))*dx`

∴ x2 + 2 = A(x + 2)(x + 3) + B(x – 1)(x + 3) + C(x – 1)(x + 2)
Put x – 1 = 0, i.e. x = 1, we get
1 + 2 = A(3)(4) + B(0)(4) + C(0)(3)

∴ 3 = 12A

∴ A = `(1)/(4)` 
Put x + 2 = 0, i.e. x = – 2, we get
4 + 2 = A(0)(1) + B(– 3)(1) + C(– 3)(0)
∴ 6 = – 3B
∴ B = – 2
Put x + 3 = 0, i.e. x = – 3we get
9 + 2 = A(– 1)(0) + B(– 4)(0) + C(– 4)(– 1)
∴ 11 = 4C

∴ C = `(11)/(4)`

∴ `(x^2 + 2)/((x - 1)(x + 2)(x + 3)) = ((1/4))/(x - 1) + (-2)/(x + 1) + ((11/4))/(x + 3)`

∴ I = `int [((1/4))/(x - 1) + (-2)/(x + 1) + ((11/4))/(x + 3)].dx`

= `(1)/(4) int (1)/(x - 1).dx - 2 int(1)/(x + 1).dx + (11)/(4) int (1)/(x + 3).dx`

`(3)/(4)log|x - 1| - log|x + 1| + (1)/(4)log|x + 3| + c`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Miscellaneous Exercise 3 [पृष्ठ १५०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Miscellaneous Exercise 3 | Q 3.17 | पृष्ठ १५०

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Integrate the rational function:

`(2x)/(x^2 + 3x + 2)`


Integrate the rational function:

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


Integrate the rational function:

`(3x + 5)/(x^3 - x^2 - x + 1)`


Integrate the rational function:

`(5x)/((x + 1)(x^2 - 4))`


Integrate the rational function:

`1/(x^4 - 1)`


Integrate the rational function:

`(cos x)/((1-sinx)(2 - sin x))` [Hint: Put sin x = t]


Integrate the rational function:

`1/(x(x^4 - 1))`


Evaluate : `∫(x+1)/((x+2)(x+3))dx`


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


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


Integrate the following with respect to the respective variable : `(cos 7x - cos8x)/(1 + 2 cos 5x)`


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


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


`int "dx"/(("x" - 8)("x" + 7))`=


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


Evaluate: `int ("3x" - 1)/("2x"^2 - "x" - 1)` dx


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


`int (2x - 7)/sqrt(4x- 1) dx`


`int ((x^2 + 2))/(x^2 + 1) "a"^(x + tan^(-1_x)) "d"x`


`int (7 + 4x + 5x^2)/(2x + 3)^(3/2) dx`


`int 1/(2 +  cosx - sinx)  "d"x`


`int sin(logx)  "d"x`


`int (x^2 + x -1)/(x^2 + x - 6)  "d"x`


`int ("d"x)/(2 + 3tanx)`


Evaluate:

`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`


Choose the correct alternative:

`int ((x^3 + 3x^2 + 3x + 1))/(x + 1)^5 "d"x` =


`int (5(x^6 + 1))/(x^2 + 1) "d"x` = x5 – ______ x3 + 5x + c


Evaluate `int x log x  "d"x`


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


Evaluate the following:

`int "e"^(-3x) cos^3x  "d"x`


The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, the denominator becomes eight times the numerator. Find the fraction.


Find: `int x^2/((x^2 + 1)(3x^2 + 4))dx`


Evaluate: `int_-2^1 sqrt(5 - 4x - x^2)dx`


If `int 1/((x^2 + 4)(x^2 + 9))dx = A tan^-1  x/2 + B tan^-1(x/3) + C`, then A – B = ______.


Evaluate:

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


Evaluate.

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


Evaluate:

`int (x + 7)/(x^2 + 4x + 7)dx`


Evaluate.

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


Evaluate:

`int(2x^3 - 1)/(x^4 + x)dx`


When is a rational function \[\frac{P(x)}{Q(x)}\] called proper?


Integration by partial fractions is a method used to integrate which functions?


Which partial form corresponds to \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x-\mathrm{b})(x-\mathrm{c})}\]?


Why is \[\frac{x^{2}+1}{x^{2}-5x+6}\] not a proper rational function?


Which factorisation is correct for \[x^{2}-5x+6\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×