English

∫ x2(x2+1)(x2-2)(x2+3) dx

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Let I = `int  x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3))  "d"x`

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

= `"A"/(x^2 + 1) + "b"/(x^2 - 2) + "c"/(x^2 + 3)`

∴ x2 = A(x2 − 2)(x2 + 3) + B(x2 + 1)(x2 + 3) + C(x2 + 1)(x2 − 2)     ........(i)

Putting x2 = 2 in (i), we get

2 = B × 3 × 5

∴ B = `2/15`

Putting x2 = −3 in (i), we get

−3 = C × (– 2) × (– 5)

∴ C = `(-3)/10`

Putting x2 = −1 in (i), we get

−1 = A × (–3) × 2

∴ A = `1/6`

∴ `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3)) = (1/6)/(x^2 + 1) + (2/15)/(x^2 - 2) + ((-3)/10)/(x^2 + 3)`

∴ I = `int[1/(6(x^2 + 1)) + 2/(15(x^2 - 2)) - 3/(10(x^2 + 3))]  "d"x`

= `1/6 int 1/(x^2 + 1)  "d"x + 2/15 int 1/(x^2 - 2)  "d"x - 3/10 int 1/(x^2 + 3)  "d"x`

= `1/6 int 1/(x^2 + 1)  "d"x + 2/15 int  1/(x^2 - (sqrt(2))^2)  "d"x - 3/10 int 1/(x^2 + (sqrt(3))^2)  "d"x`

= `1/6 tan^-1x + 2/15 xx 1/(2 xx sqrt(2)) log|(x - sqrt(2))/(x + sqrt(2))| - 3/10 xx 1/sqrt(3) tan^-1 (x/sqrt(3)) + "c"`

∴ I = `1/6 tan^-1x + 1/(15sqrt(2)) log|(x - sqrt(2))/(x + sqrt(2))| - sqrt(3)/10 tan^-1 (x/sqrt(3)) + "c"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Long Answers III

APPEARS IN

RELATED QUESTIONS

Evaluate: `∫8/((x+2)(x^2+4))dx` 


Integrate the rational function:

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


Integrate the rational function:

`(3x - 1)/((x - 1)(x - 2)(x - 3))`


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

`((x^2 +1)(x^2 + 2))/((x^2 + 3)(x^2+ 4))`


Integrate the rational function:

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


Integrate the rational function:

`1/(e^x -1)`[Hint: Put ex = t]


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


Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`


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


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


Integrate the following w.r.t. x:

`(6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1)`


Integrate the following w.r.t. x : `(1)/(x(1 + 4x^3 + 3x^6)`


Integrate the following w.r.t.x :  `sec^2x sqrt(7 + 2 tan x - tan^2 x)`


Integrate the following w.r.t.x : `sqrt(tanx)/(sinx*cosx)`


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


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


`int x^2sqrt("a"^2 - x^6)  "d"x`


`int (sinx)/(sin3x)  "d"x`


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


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


`int x^3tan^(-1)x  "d"x`


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


`int (3"e"^(2x) + 5)/(4"e"^(2x) - 5)  "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 = ?


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


Evaluate `int (2"e"^x + 5)/(2"e"^x + 1)  "d"x`


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


Verify the following using the concept of integration as an antiderivative

`int (x^3"d"x)/(x + 1) = x - x^2/2 + x^3/3 - log|x + 1| + "C"`


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


Value of ∫ `(x^2 + 1)/((x − 1)(x − 2))`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×