हिंदी

Evaluate: int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

मूल्यांकन
Advertisements

उत्तर

Consider, `(2x^2 - 3)/((x^2 - 5)(x^2 + 4))`

Let x2 = m

∴ `(2m - 3)/((m - 5)(m + 4))`  ...[proper rational function]

Now, `(2m - 3)/((m - 5)(m + 4)) = A/((m - 5)) + B/((m + 4))`

`(2m - 3)/((m - 5)(m + 4)) = (A(m + 4) + B(m - 5))/((m - 5)(m + 4))`

∴ 2m – 3 = A (m + 4) + B (m – 5)

at m = 5, 2(5) – 3 = A (9) + B (0)

7 = 9A `\implies` A = `7/9`

at m = –4, 2(–4) – 3 = A (0) + B (–9)

–11 = –9B `\implies` B = `11/9`

Thus, `(2m - 3)/((m - 5)(m + 4)) = ((7/9))/((m - 5)) + ((11/9))/((m + 4))` i.e. `(2x^2 - 3)/((x^2 - 5)(x^2 + 4)) = ((7/9))/((x^2 - 5)) + ((11/9))/((x^2 + 4))`

∴ I = `int [((7/9))/(x^2 - 5) + ((11/9))/(x^2 + 4)]dx`

= `7/9 . int 1/(x^2 - (sqrt(5))^2)dx + 11/9 . int 1/(x^2 + (2)^2)dx`

= `7/9 . 1/(2(sqrt(5))) . log [(x - sqrt(5))/(x + sqrt(5))] + 11/9 . 1/2 . tan^-1 (x/2) + c`

∴ I = `7/(18(sqrt(5))) . log [(x - sqrt(5))/(x + sqrt(5))] + 11/18 . tan^-1 (x/2) + c`

∴ `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx = 7/(18(sqrt5)) . log [(x - sqrt(5))/(x + sqrt(5))] + 11/18 . tan^-1 (x/2) + c`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2022-2023 (March) Official

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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


Find: `I=intdx/(sinx+sin2x)`


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

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


Integrate the rational function:

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


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:

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


Find : 

`∫ sin(x-a)/sin(x+a)dx`


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


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


Evaluate:

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


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


State whether the following statement is True or False.

If `int (("x - 1") "dx")/(("x + 1")("x - 2"))` = A log |x + 1| + B log |x - 2| + c, then A + B = 1.


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


`int 1/(4x^2 - 20x + 17)  "d"x`


`int sec^3x  "d"x`


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


`int x^3tan^(-1)x  "d"x`


`int x sin2x cos5x  "d"x`


`int (x + sinx)/(1 - cosx)  "d"x`


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


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


`int xcos^3x  "d"x`


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


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


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


State whether the following statement is True or False:

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


Evaluate `int x^2"e"^(4x)  "d"x`


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


Evaluate the following:

`int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`


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 ______.


If `int dx/sqrt(16 - 9x^2)` = A sin–1 (Bx) + C then A + B = ______.


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


Evaluate:

`int x/((x + 2)(x - 1)^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×