Advertisements
Advertisements
प्रश्न
`int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
Advertisements
उत्तर
Let I = `int (6x^2 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
2x + 3
`3x^2 - 2x - 1")"overline(6x^3 + 5x^2 + 0x - 7`
6x3 − 4x2 − 2x
(−) (+) (+)
9x2 + 2x − 7
9x2 − 6x − 3
(−) (+) (+)
8x − 4
∴ I = `int (2x + 3 + (8x - 4)/(3x^2 - 2x - 1)) "d"x`
3x2 – 2x – 1 = 3x2 – 3x + x – 1
= 3x(x – 1) + 1(x – 1)
= (x – 1)(3 x + 1)
∴ I = `int[2x + 3 + (8x - 4)/((x - 1)(3x + 1))] "d"x`
Let `(8x - 4)/((x - 1)(3x + 1)) = "A"/(x - 1) + "B"/(3x + 1)`
∴ 8x – 4 = A(3x + 1) + B(x – 1) ........(i)
Putting x = 1 in (i), we get
4 = 4A
∴ A = 1
Putting x = `(-1)/3` in (i), we get
`8(-1/3) - 4 = "B"(-1/3 - 1)`
∴ `(-20)/3 = -4/3 "B"`
∴ B = 5
∴ `(8x - 4)/((x - 1)(3x + 1)) = 1/(x - 1) + 5/(3x + 1)`
∴ I = `int (2x + 3 + 1/(x - 1) + 5/(3x + 1)) "d"x`
= `2 int x "d"x + 3 int "d"x + int 1/(x - 1) "d"x + 5/3 int 3/(3x + 1) "d"x`
= `2(x^2/2) + 3x + log|x + 1| + (5log|3x + 1|)/3 + "c"`
∴ I = `x^2 + 3x + log|x - 1| + 5/3 log|3x + 1| + "c"`
APPEARS IN
संबंधित प्रश्न
Find : `int x^2/(x^4+x^2-2) dx`
Integrate the rational function:
`x/((x + 1)(x+ 2))`
Integrate the rational function:
`(1 - x^2)/(x(1-2x))`
Integrate the rational function:
`x/((x^2+1)(x - 1))`
Integrate the rational function:
`x/((x -1)^2 (x+ 2))`
Integrate the rational function:
`(5x)/((x + 1)(x^2 - 4))`
Integrate the rational function:
`1/(x(x^4 - 1))`
Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`
Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`
Integrate the following w.r.t. x : `(5x^2 + 20x + 6)/(x^3 + 2x ^2 + x)`
Integrate the following w.r.t. x : `(1)/(x(1 + 4x^3 + 3x^6)`
Integrate the following w.r.t. x: `(1)/(sinx + sin2x)`
Integrate the following w.r.t. x : `(1)/(sinx*(3 + 2cosx)`
Integrate the following w.r.t. x: `(2x^2 - 1)/(x^4 + 9x^2 + 20)`
Integrate the following with respect to the respective variable : `(cos 7x - cos8x)/(1 + 2 cos 5x)`
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 : `sqrt(tanx)/(sinx*cosx)`
Evaluate: `int (2"x" + 1)/(("x + 1")("x - 2"))` dx
Evaluate:
`int (2x + 1)/(x(x - 1)(x - 4)) 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.
Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
`int (2x - 7)/sqrt(4x- 1) dx`
`int x^7/(1 + x^4)^2 "d"x`
`int x^2sqrt("a"^2 - x^6) "d"x`
If f'(x) = `x - 3/x^3`, f(1) = `11/2` find f(x)
`int "e"^x ((1 + x^2))/(1 + x)^2 "d"x`
`int (3x + 4)/sqrt(2x^2 + 2x + 1) "d"x`
`int x sin2x cos5x "d"x`
`int 1/(sinx(3 + 2cosx)) "d"x`
`int (sin2x)/(3sin^4x - 4sin^2x + 1) "d"x`
`int (3"e"^(2x) + 5)/(4"e"^(2x) - 5) "d"x`
Choose the correct alternative:
`int sqrt(1 + x) "d"x` =
If f'(x) = `1/x + x` and f(1) = `5/2`, then f(x) = log x + `x^2/2` + ______ + c
`int 1/x^3 [log x^x]^2 "d"x` = p(log x)3 + c Then p = ______
If `int(sin2x)/(sin5x sin3x)dx = 1/3log|sin 3x| - 1/5log|f(x)| + c`, then f(x) = ______
Verify the following using the concept of integration as an antiderivative
`int (x^3"d"x)/(x + 1) = x - x^2/2 + x^3/3 - log|x + 1| + "C"`
Evaluate the following:
`int x^2/(1 - x^4) "d"x` put x2 = t
Evaluate the following:
`int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`
Evaluate the following:
`int sqrt(tanx) "d"x` (Hint: Put tanx = t2)
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.
Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`
Find : `int (2x^2 + 3)/(x^2(x^2 + 9))dx; x ≠ 0`.
Evaluate`int(5x^2-6x+3)/(2x-3)dx`
Evaluate:
`int x/((x + 2)(x - 1)^2)dx`
