English

∫ x2(x2+1)(x2-2)(x2+3) dx - Mathematics and Statistics

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

Find: `I=intdx/(sinx+sin2x)`


Evaluate: `∫8/((x+2)(x^2+4))dx` 


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

`1/(x(x^n + 1))` [Hint: multiply numerator and denominator by xn − 1 and put xn = t]


Integrate the rational function:

`(2x)/((x^2 + 1)(x^2 + 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 : `(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 : `(1)/(x(1 + 4x^3 + 3x^6)`


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


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


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


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


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


Choose the correct options from the given alternatives :

If `int tan^3x*sec^3x*dx = (1/m)sec^mx - (1/n)sec^n x + c, "then" (m, n)` =


Integrate the following with respect to the respective variable : `(6x + 5)^(3/2)`


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


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


Integrate the following w.r.t.x : `(1)/((1 - cos4x)(3 - cot2x)`


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


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


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


`int x^7/(1 + x^4)^2  "d"x`


`int sqrt(4^x(4^x + 4))  "d"x`


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


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


`int sin(logx)  "d"x`


`int x sin2x cos5x  "d"x`


`int 1/(sinx(3 + 2cosx))  "d"x`


`int xcos^3x  "d"x`


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


Choose the correct alternative:

`int sqrt(1 + x)  "d"x` =


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


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 `int x^2"e"^(4x)  "d"x`


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


`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5)  "dt"`


Evaluate the following:

`int (x^2"d"x)/(x^4 - x^2 - 12)`


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


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


Evaluate:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×