Advertisements
Advertisements
प्रश्न
If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then ______.
पर्याय
a = `(-1)/10`, b = `(-2)/5`
a = `1/10`, b = `- 2/5`
a = `(-1)/10`, b = `2/5`
a = `1/10`, b = `2/5`
Advertisements
उत्तर
If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then a = `(-1)/10`, b = `2/5`.
Explanation:
Given that, `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`
Now, I = `int "dx"/((x + 2)(x^2 + 1))`
`1/((x + 2)(x^2 + 1)) = "A"/(x + 2) + ("B"x + "C")/(x^2 + 1)`
⇒ 1 = A(x2 + 1) + (Bx + C)(x + 2)
⇒ 1 = (A + B)x2 + (2B + C)x + A + 2C
Comapring coefficient, we get
A + B = 0
A + 2C = 1
2B + C = 0
Solving we get A = `1/5`
B = `- 1/5`
And C = `2/5`
∴ `int "dx"/((x + 2)(x^2 + 1))`
= `1/5 int 1/(x + 2) "d"x + int (- 1/5 + 2/5)/(x^2 + 1) "d"x`
= `1/5 int 1/(x + 2) "d"x - 1/10 int (2x)/(1 + x^2) "d"x + 1/5 int 2/(1 + x^2) "d"x`
= `1/5 log|x + 2| - 1/10 log|1 + x^2| + 2/5 tan^-1x + "C"`
= `"a" log |1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"` ....(Given)
∴ a = `(-1)/10`, b = `2/5`.
APPEARS IN
संबंधित प्रश्न
Evaluate : `int x^2/((x^2+2)(2x^2+1))dx`
Evaluate:
`int x^2/(x^4+x^2-2)dx`
Find: `I=intdx/(sinx+sin2x)`
Integrate the rational function:
`x/((x + 1)(x+ 2))`
Integrate the rational function:
`x/((x-1)(x- 2)(x - 3))`
Integrate the rational function:
`(3x + 5)/(x^3 - x^2 - x + 1)`
Integrate the rational function:
`(x^3 + x + 1)/(x^2 -1)`
Integrate the rational function:
`1/(x^4 - 1)`
Integrate the rational function:
`1/(e^x -1)`[Hint: Put ex = t]
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 : `(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 : `(2log x + 3)/(x(3 log x + 2)[(logx)^2 + 1]`
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/((x - 1)^2(x + 2)) dx`
For `int ("x - 1")/("x + 1")^3 "e"^"x" "dx" = "e"^"x"` f(x) + c, f(x) = (x + 1)2.
Evaluate: `int (1 + log "x")/("x"(3 + log "x")(2 + 3 log "x"))` dx
`int (2x - 7)/sqrt(4x- 1) dx`
`int x^7/(1 + x^4)^2 "d"x`
`int sec^2x sqrt(tan^2x + tanx - 7) "d"x`
`int (x^2 + x -1)/(x^2 + x - 6) "d"x`
`int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
`int (x + sinx)/(1 - cosx) "d"x`
`int (sin2x)/(3sin^4x - 4sin^2x + 1) "d"x`
Choose the correct alternative:
`int sqrt(1 + x) "d"x` =
Choose the correct alternative:
`int ((x^3 + 3x^2 + 3x + 1))/(x + 1)^5 "d"x` =
`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5) "dt"`
Evaluate the following:
`int (x^2"d"x)/(x^4 - x^2 - 12)`
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 `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 = ______.
Find : `int (2x^2 + 3)/(x^2(x^2 + 9))dx; x ≠ 0`.
Evaluate`int(5x^2-6x+3)/(2x-3)dx`
