English

∫x(x-1)2(x+2)dx - Mathematics and Statistics

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

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

Let `x/((x - 1)^2 (x + 2)) = "A"/(x - 1) + "B"/(x - 1)^2 + "C"/((x + 2))`

∴ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)2    ......(i)

Putting x = 1 in (i), we get

1 = A(0)(3) + B(3) + C(0)2

∴ 1 = 3B

∴ B = `1/3`

Putting x =  2 in (i), we get

– 2 = A(– 3)(0) + B(0) + C(9)

∴ – 2 = 9C

∴ C = `-2/9`

Putting x = – 1 in (i), we get

– 1 = A(– 2)(1) + B(1) + C(4)

∴ – 1 = `-2"A" + 1/3 - 8/9` 

∴ – 1 = `-2"A" - 5/9`

∴ 2A = `-5/9 + 1 = 4/9`

∴ A = `2/9`

∴ `x/((x - 1)^2(x + 2)) = (2/9)/(x - 1) + (1/3)/(x - 1)^2 + ((-2/9))/(x + 2)`

∴ I = `int[(2/9)/(x - 1) + (1/3)/(x - 1)^2 + ((-2/9))/(x + 2)] "d"x`

= `2/9 int 1/(x - 1) "d"x + 1/3int(x - 1)^(-2) "d"x - 2/9 int 1/(x + 2) "d"x`

= `2/9 log|x - 1| + 1/3*((x - 1)^(-1))/(-1) - 2/9 log|x + 2| + "c"`

= `2/9 log|x - 1| - 2/9 log|x + 2| - 1/3 xx 1/((x - 1)) + "c"`

∴ I = `2/9 log|(x - 1)/(x + 2)| - 1/(3(x - 1)) + "c"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 1.5: Integration - Q.5

RELATED QUESTIONS

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:

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


Integrate the rational function:

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


Integrate the rational function:

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


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


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


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)/(sinx*(3 + 2cosx)`


Integrate the following w.r.t.x:

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


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


`int x^7/(1 + x^4)^2  "d"x`


`int sqrt(4^x(4^x + 4))  "d"x`


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


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


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


`int  ((2logx + 3))/(x(3logx + 2)[(logx)^2 + 1])  "d"x`


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


Evaluate `int x^2"e"^(4x)  "d"x`


Evaluate the following:

`int_"0"^pi  (x"d"x)/(1 + sin x)`


Evaluate the following:

`int "e"^(-3x) cos^3x  "d"x`


Let g : (0, ∞) `rightarrow` R be a differentiable function such that `int((x(cosx - sinx))/(e^x + 1) + (g(x)(e^x + 1 - xe^x))/(e^x + 1)^2)dx = (xg(x))/(e^x + 1) + c`, for all x > 0, where c is an arbitrary constant. Then ______.


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


Evaluate: 

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


Evaluate.

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


If \[\int\frac{2x+3}{(x-1)(x^{2}+1)}\mathrm{d}x\] = \[=\log_{e}\left\{(x-1)^{\frac{5}{2}}\left(x^{2}+1\right)^{a}\right\}-\frac{1}{2}\tan^{-1}x+\mathrm{A}\] where A is an arbitrary constant, then the value of a is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×