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 = ?
पर्याय
`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
Integrate the functions:
`(log x)^2/x`
Integrate the functions:
`1/(1 - tan x)`
`(10x^9 + 10^x log_e 10)/(x^10 + 10^x) dx` equals:
Evaluate `int 1/(3+ 2 sinx + cosx) dx`
Write a value of\[\int a^x e^x \text{ dx }\]
Evaluate the following integrals : `int (sin2x)/(cosx)dx`
Evaluate the following integrals:
`int x/(x + 2).dx`
Evaluate the following integrals : `int (3)/(sqrt(7x - 2) - sqrt(7x - 5)).dx`
Integrate the following functions w.r.t. x : `((x - 1)^2)/(x^2 + 1)^2`
Evaluate the following : `int (1)/(4x^2 - 3).dx`
Evaluate the following : `int sqrt((10 + x)/(10 - x)).dx`
Evaluate the following : `int (1)/(cos2x + 3sin^2x).dx`
Evaluate the following.
`int ("e"^"x" + "e"^(- "x"))^2 ("e"^"x" - "e"^(-"x"))`dx
`int x^2/sqrt(1 - x^6)` dx = ________________
`int 2/(sqrtx - sqrt(x + 3))` dx = ________________
`int sin^-1 x`dx = ?
`int (x^2 + 1)/(x^4 - x^2 + 1)`dx = ?
If `tan^-1x = 2tan^-1((1 - x)/(1 + x))`, then the value of x is ______
The general solution of the differential equation `(1 + y/x) + ("d"y)/(d"x)` = 0 is ______.
Prove that:
`int 1/sqrt(x^2 - a^2) dx = log |x + sqrt(x^2 - a^2)| + c`.
Evaluate.
`int (5x^2 - 6x + 3)/(2x - 3) dx`
Evaluate.
`int (5x^2-6x+3)/(2x-3)dx`
Evaluate `int 1/(x(x-1))dx`
Evaluate `int1/(x(x-1))dx`
`int (x + 1)/(x(1 + xe^x)) dx` is equal to
For \[\int\frac{\sin x}{\sin(x+a)}\,dx\], which substitution gives \[dx=dt\]?
Integration by substitution is the reverse process of which rule?
