Advertisements
Advertisements
प्रश्न
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
Advertisements
उत्तर
Let I = `int (2x + 1)/((x + 1)(x - 2)) "d"x`
Let `(2x + 1)/((x + 1)(x - 2)) = "A"/(x + 1) + "B"/(x - 2)`
∴ 2x + 1 = A(x – 2) + B(x + 1) ......(i)
Putting x = – 1 in (i), we get
2(– 1) + 1 = A(– 1 – 2) + B(0)
∴ – 1 = – 3A
∴ A = `1/3`
Putting x = 2 in (i), we get
2(2) + 1 = A(0) + B(2 + 1)
∴ 5 = 3B
∴ B = `5/3`
∴ `(2x + 1)/((x + 1)(x - 2)) = ((1/3))/(x + 1) + ((5/3))/(x - 2)`
∴ I = `int(((1/3))/(x + 1) + ((5/3))/(x - 2)) "d"x`
= `1/3 int 1/(x + 1) "d"x + 5/3 int 1/(x - 2) "d"x`
∴ I = `1/3 log|x + 1| + 5/3 log|x - 2| + "c"`
संबंधित प्रश्न
Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`
Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`
Evaluate the following : `int cos sqrt(x).dx`
Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`
Integrate the following functions w.r.t. x: `sqrt(x^2 + 2x + 5)`.
Choose the correct options from the given alternatives :
`int (x- sinx)/(1 - cosx)*dx` =
Integrate the following w.r.t.x : sec4x cosec2x
Evaluate the following.
`int x^2 e^4x`dx
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
Evaluate the following.
`int (log "x")/(1 + log "x")^2` dx
`int ("x" + 1/"x")^3 "dx"` = ______
Evaluate:
∫ (log x)2 dx
`int 1/(4x + 5x^(-11)) "d"x`
`int sin4x cos3x "d"x`
∫ log x · (log x + 2) dx = ?
`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.
Find `int_0^1 x(tan^-1x) "d"x`
Solve: `int sqrt(4x^2 + 5)dx`
If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.
Find `int e^(cot^-1x) ((1 - x + x^2)/(1 + x^2))dx`.
Solution of the equation `xdy/dx=y log y` is ______
Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.
Solution: (x2 + y2) dx - 2xy dy = 0
∴ `dy/dx=(x^2+y^2)/(2xy)` ...(1)
Puty = vx
∴ `dy/dx=square`
∴ equation (1) becomes
`x(dv)/dx = square`
∴ `square dv = dx/x`
On integrating, we get
`int(2v)/(1-v^2) dv =intdx/x`
∴ `-log|1-v^2|=log|x|+c_1`
∴ `log|x| + log|1-v^2|=logc ...["where" - c_1 = log c]`
∴ x(1 - v2) = c
By putting the value of v, the general solution of the D.E. is `square`= cx
Evaluate:
`int e^(logcosx)dx`
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3) dx`
