Advertisements
Advertisements
प्रश्न
Integrate the following w.r.t. x : `(5x^2 + 20x + 6)/(x^3 + 2x ^2 + x)`
Advertisements
उत्तर
Let I = `int (5x^2 + 20x + 6)/(x^3 + 2x^2 + x)*dx`
= `int (5x^2 + 20x + 6)/(x(x^2 + 2x + 1))*dx`
= `int (5x^2 + 20x + 6)/(x(x + 1)^2)*dx`
Let `(5x^2 + 20x + 6)/(x(x + 1)^2) = "A"/x + "B"/(x + 1) + "C"/(x + 1)^2`
∴ 5x2 + 20x + 6 = A(x + 1)2 + Bx(x + 1) + Cx
Put x = 0, we get
0 + 0 + 6 = A(1) + B(0)(1) + C(0)
∴ A = 6
Put x + 1 = 0, i x = – 1, we get
5(1) + 20(– 1) + 6 = A(0) + B(– 1)(0) + C(– 1)
∴ – 9 = – C
∴ C = 9
Put x = 1, we get
5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1)
But A = 6 and C = 9
∴ 31 = 24 + 2B + 9
∴ B = – 1
∴ `(5x^2 + 20x + 6)/(x(x + 1)^2) = 6/x - (1)/(x + 1) + (9)/(x + 1)^2`
∴ I = `int[6/x - (1)/(x + 1) + 9/(x + 1)^2]*dx`
= `6 int 1/x*dx - int 1/(x + 1)*dx + 9 int (x + 1)^-2*dx`
= `6log|x| - log|x + 1| + 9*((x + 1)^-1)/(-1) + c`
= `log|x^6| - log|x + 1| - (9)/((x + 1)) + c`
= `log|x^6/(x + 1)| - (9)/((x + 1)) + c`.
APPEARS IN
संबंधित प्रश्न
Find : `int x^2/(x^4+x^2-2) dx`
Evaluate:
`int x^2/(x^4+x^2-2)dx`
Integrate the rational function:
`(1 - x^2)/(x(1-2x))`
Integrate the rational function:
`(3x -1)/(x + 2)^2`
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))`
`int (xdx)/((x - 1)(x - 2))` equals:
Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`
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 : `(12x + 3)/(6x^2 + 13x - 63)`
Integrate the following w.r.t. x : `(2x)/(4 - 3x - x^2)`
Integrate the following w.r.t. x : `(2x)/((2 + x^2)(3 + 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)/(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 : `(5*e^x)/((e^x + 1)(e^(2x) + 9)`
Integrate the following w.r.t. x: `(x^2 + 3)/((x^2 - 1)(x^2 - 2)`
Integrate the following with respect to the respective variable : `(cos 7x - cos8x)/(1 + 2 cos 5x)`
Integrate the following w.r.t.x : `sec^2x sqrt(7 + 2 tan x - tan^2 x)`
Integrate the following w.r.t.x: `(x + 5)/(x^3 + 3x^2 - x - 3)`
Evaluate: `int 1/("x"("x"^"n" + 1))` dx
Evaluate: `int ("3x" - 1)/("2x"^2 - "x" - 1)` dx
`int (2x - 7)/sqrt(4x- 1) dx`
`int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
`int "e"^(sin^(-1_x))[(x + sqrt(1 - x^2))/sqrt(1 - x^2)] "d"x`
`int "e"^x ((1 + x^2))/(1 + x)^2 "d"x`
`int ("d"x)/(x^3 - 1)`
Evaluate:
`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`
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
`int x/((x - 1)^2 (x + 2)) "d"x`
Evaluate the following:
`int x^2/(1 - x^4) "d"x` put x2 = t
Evaluate the following:
`int_"0"^pi (x"d"x)/(1 + sin x)`
The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, the denominator becomes eight times the numerator. Find the fraction.
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)
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 ______.
`int 1/(x^2 + 1)^2 dx` = ______.
Find: `int x^4/((x - 1)(x^2 + 1))dx`.
Evaluate:
`int (x + 7)/(x^2 + 4x + 7)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
