Advertisements
Advertisements
प्रश्न
`int ("x + 2")/(2"x"^2 + 6"x" + 5)"dx" = "p" int (4"x" + 6)/(2"x"^2 + 6"x" + 5) "dx" + 1/2 int "dx"/(2"x"^2 + 6"x" + 5)`, then p = ?
विकल्प
`1/3`
`1/2`
`1/4`
2
Advertisements
उत्तर
`1/4`
Explanation:
Let x + 2 = p `"d"/"dx" (2"x"^2 + 6"x" + 5) + "q"`
= p(4x + 6) + q
∴ x + 2 = 4px + 6p + q
∴ 4p = 1 and 6p + q = 2
∴ p = `1/4`
APPEARS IN
संबंधित प्रश्न
Evaluate : `∫1/(cos^4x+sin^4x)dx`
Integrate the functions:
`sin x/(1+ cos x)`
Write a value of
Write a value of
Write a value of
Write a value of\[\int\sqrt{9 + x^2} \text{ dx }\].
Integrate the following w.r.t. x : `(3x^3 - 2x + 5)/(xsqrt(x)`
Evaluate the following integrals:
`int x/(x + 2).dx`
Evaluate the following integrals:
`int (sin4x)/(cos2x).dx`
Integrate the following functions w.r.t.x:
`(5 - 3x)(2 - 3x)^(-1/2)`
Integrate the following functions w.r.t. x : `x^2/sqrt(9 - x^6)`
Integrate the following functions w.r.t. x : `(cos3x - cos4x)/(sin3x + sin4x)`
Integrate the following functions w.r.t. x : tan5x
Evaluate `int (-2)/(sqrt("5x" - 4) - sqrt("5x" - 2))`dx
Evaluate the following.
`int "x"^3/(16"x"^8 - 25)` dx
Evaluate `int (5"x" + 1)^(4/9)` dx
Evaluate: `int "e"^sqrt"x"` dx
`int 2/(sqrtx - sqrt(x + 3))` dx = ________________
`int (2(cos^2 x - sin^2 x))/(cos^2 x + sin^2 x)` dx = ______________
`int x^x (1 + logx) "d"x`
`int (x^2 + 1)/(x^4 - x^2 + 1)`dx = ?
If `int x^3"e"^(x^2) "d"x = "e"^(x^2)/2 "f"(x) + "c"`, then f(x) = ______.
`int(7x - 2)^2dx = (7x -2)^3/21 + c`
`int 1/(sinx.cos^2x)dx` = ______.
Evaluate `int(1+ x + x^2/(2!)) dx`
Evaluate the following.
`int(20 - 12"e"^"x")/(3"e"^"x" - 4) "dx"`
if `f(x) = 4x^3 - 3x^2 + 2x +k, f (0) = - 1 and f (1) = 4, "find " f(x)`
Evaluate:
`int(sqrt(tanx) + sqrt(cotx))dx`
Evaluate.
`int (5x^2 -6x + 3)/(2x -3)dx`
Evaluate `int (1 + x + x^2/(2!)) dx`
