Advertisements
Advertisements
प्रश्न
Integrate the rational function:
`((x^2 +1)(x^2 + 2))/((x^2 + 3)(x^2+ 4))`
Advertisements
उत्तर
`((x^2 + 1)(x^2 + 2))/((x^2 + 3)(x^2 + 4))` Taking x2 = y
`((y + 1)(y + 2))/((y + 3)(y + 4)) = (y^2 + 3y + 2)/(y^2 + 7y + 12)`
`= 1 - (4y + 10)/(y^2 + 7y + 12)`
`= 1 - (4y + 10)/((y + 3)(y + 4))`
Let `(4y + 10)/((y + 3)(y + 4)) = A/((y + 3)) + B/(y + 4)`
4y + 10 = A (y + 4) + B (y + 3)
Putting y = -4 - 6 = 0 - B
⇒ B = 6
Putting y = -3, -2 = A + 0
⇒ A = -2
`therefore ((x^2 + 1)(x^2 + 2))/((x^2 + 3)(x^2 + 4)) = 1 - [(-2)/(y + 3) + 6/(y + 4)]`
`= 1 + 2/(y + 3) + 6/(y + 4)`
`int ((x^2 + 1)(x^2 + 2))/((x^2 + 3)(x^2 + 4))` dx
`= int dx + 2 int 1/(x^2 sqrt(3^2)) + 6 int 1/(x^2 + 4)` dx
`= x + 2/sqrt 3 tan^-1 x/sqrt3 - 6/2 tan^-1 (x/2) + C`
`= x + 2/sqrt 3 tan^-1 x/sqrt3 - 3 tan^-1 x/2 + C`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int x^2/((x^2+2)(2x^2+1))dx`
Integrate the rational function:
`1/(x^2 - 9)`
Integrate the rational function:
`(3x + 5)/(x^3 - x^2 - x + 1)`
Integrate the rational function:
`2/((1-x)(1+x^2))`
Integrate the rational function:
`1/(x^4 - 1)`
Integrate the rational function:
`(cos x)/((1-sinx)(2 - sin x))` [Hint: Put sin x = t]
Evaluate : `∫(x+1)/((x+2)(x+3))dx`
Find `int(e^x dx)/((e^x - 1)^2 (e^x + 2))`
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)/(2sinx + sin2x)`
Integrate the following with respect to the respective variable : `(6x + 5)^(3/2)`
Integrate the following with respect to the respective variable : `cot^-1 ((1 + sinx)/cosx)`
Integrate the following w.r.t.x : `x^2/sqrt(1 - x^6)`
Integrate the following w.r.t.x : `(1)/(sinx + sin2x)`
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 (2x + 1)/(x(x - 1)(x - 4)) 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
Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
`int x^7/(1 + x^4)^2 "d"x`
`int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
`int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
Choose the correct alternative:
`int ((x^3 + 3x^2 + 3x + 1))/(x + 1)^5 "d"x` =
If f'(x) = `1/x + x` and f(1) = `5/2`, then f(x) = log x + `x^2/2` + ______ + c
Evaluate `int (2"e"^x + 5)/(2"e"^x + 1) "d"x`
Evaluate the following:
`int_"0"^pi (x"d"x)/(1 + sin x)`
Evaluate: `int (dx)/(2 + cos x - sin x)`
If `int dx/sqrt(16 - 9x^2)` = A sin–1 (Bx) + C then A + B = ______.
Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`
Evaluate`int(5x^2-6x+3)/(2x-3)dx`
Evaluate:
`int 2/((1 - x)(1 + x^2))dx`
Evaluate.
`int (5x^2 - 6x + 3)/(2x - 3)dx`
Evaluate:
`int(2x^3 - 1)/(x^4 + x)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
