Advertisements
Advertisements
प्रश्न
`int ("d"x)/(x^3 - 1)`
Advertisements
उत्तर
Let I = `int ("d"x)/(x^3 - 1)`
= `int 1/((x - 1)(x^2 + x + 1)) "d"x`
Let `1/((x - 1)(x^2 + x + 1))`
= `"A"/(x - 1) + ("B"x + "C")/(x^2 + x + 1)`
∴ 1 = A(x2 + x + 1) + (Bx + C)(x – 1) .......(i)
Putting x = 1 in (i), we get
1 = A(12 + 1 + 1)
∴ 1 = 3A
∴ A = `1/3`
Putting x = 0 in (i), we get
1 = A(0 + 0 + 1) + (0 + C)(0 – 1)
∴ 1 = A – C
∴ 1 = `1/3 - "C"`
∴ C = `- 2/3`
Putting x = 2 in (i), we get
1 = A(22 + 2 + 1) + (2B + C)(2 – 1)
∴ 1 = 7A + 2B + C
∴ 1 = `7/3 + 2"B" - 2/3`
∴ 1 = `5/3 + 2"B"`
∴ `(-2)/(3)` = 2B
∴ B = `-1/3`
∴ I = `int ((1/3)/(x - 1) + (-1/3x - 2/3)/(x^2 + x + 1)) "d"x`
= `1/3 int(1/(x - 1) - (x + 2)/(x^2 + x + 1)) "d"x`
= `1/3 int 1/(x - 1) "d"x - 1/3 int (x + 2)/(x^2 + x + 1) "d"x`
= `1/3 int 1/(x - 1) "d"x - 1/3*1/2 int (2x + 4)/(x^2 + x + 1) "d"x`
= `1/3 int 1/(x 1) "d"x - 1/6 int ((2x + 1) + 3)/(x^2 + x + 1)* "d"x`
= `1/3 int 1/(x - 1) "d"x - 1/6 int (2x + 1)/(x^2 + x + 1) "d"x - 1/2 int ("d"x)/(x^2 + x + 1)`
= `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/2 int ("d"x)/(x^2 + x + 1/4 - 1/4 + 1)` ......`[∵ int ("f'"(x))/("f"(x)) "d"x = log|"f"(x)| + "c"]`
= `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/2 int ("d"x)/((x + 1/2)^2 + (sqrt(3)/2)^2`
= `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/2* 1/(sqrt(3)/2) tan^-1 ((x + 1/2)/(sqrt(3)/2)) + "c"`
∴ I = `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/sqrt(3) tan^-1 ((2x + 1)/sqrt(3)) + "c"`
APPEARS IN
संबंधित प्रश्न
Find : `int x^2/(x^4+x^2-2) dx`
Evaluate: `∫8/((x+2)(x^2+4))dx`
Integrate the rational function:
`(x^3 + x + 1)/(x^2 -1)`
Integrate the rational function:
`(cos x)/((1-sinx)(2 - sin x))` [Hint: Put sin x = t]
Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`
Integrate the following w.r.t. x : `(x^2 + 2)/((x - 1)(x + 2)(x + 3)`
Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`
Integrate the following w.r.t. x:
`(6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1)`
Integrate the following w.r.t. x : `(1)/(x(x^5 + 1)`
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)/(2sinx + sin2x)`
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 : `sqrt(tanx)/(sinx*cosx)`
Evaluate: `int ("x"^2 + "x" - 1)/("x"^2 + "x" - 6)` dx
Evaluate:
`int x/((x - 1)^2(x + 2)) dx`
Evaluate: `int 1/("x"("x"^5 + 1))` dx
For `int ("x - 1")/("x + 1")^3 "e"^"x" "dx" = "e"^"x"` f(x) + c, f(x) = (x + 1)2.
Evaluate: `int ("3x" - 1)/("2x"^2 - "x" - 1)` dx
`int 1/(x(x^3 - 1)) "d"x`
`int sqrt((9 + x)/(9 - x)) "d"x`
`int sec^3x "d"x`
`int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
`int (x^2 + x -1)/(x^2 + x - 6) "d"x`
`int x sin2x cos5x "d"x`
`int (x + sinx)/(1 - cosx) "d"x`
`int xcos^3x "d"x`
`int (sin2x)/(3sin^4x - 4sin^2x + 1) "d"x`
`int 1/x^3 [log x^x]^2 "d"x` = p(log x)3 + c Then p = ______
`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5) "dt"`
If `int(sin2x)/(sin5x sin3x)dx = 1/3log|sin 3x| - 1/5log|f(x)| + c`, then f(x) = ______
Evaluate the following:
`int (x^2"d"x)/(x^4 - x^2 - 12)`
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 = ______.
If `int 1/((x^2 + 4)(x^2 + 9))dx = A tan^-1 x/2 + B tan^-1(x/3) + C`, then A – B = ______.
If `intsqrt((x - 5)/(x - 7))dx = Asqrt(x^2 - 12x + 35) + log|x| - 6 + sqrt(x^2 - 12x + 35) + C|`, then A = ______.
Find: `int x^4/((x - 1)(x^2 + 1))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
