Advertisements
Advertisements
प्रश्न
Integrate the function in `(xe^x)/(1+x)^2`.
Advertisements
उत्तर
Let `I = int (xe^x)/((1 + x)^2) dx`
`= int ((x + 1 - 1) e^x)/((1 + x)^2) dx`
`= int 1/((1 + x)) . e^x dx - (e^x - 1)/((1 + x)^2) dx`
`= 1/((1 + x)). e^x - int (-1)/((1 + x^2)).e^x dx - int e^x/((1 + x^2)) dx + C`
`= e^x/(1 + x) + int e^x/((1 + x)^2) dx - int e^x/((1 + x)^2) dx + C`
`= e^x/(1 + x) + C`
APPEARS IN
संबंधित प्रश्न
Integrate : sec3 x w. r. t. x.
`int1/xlogxdx=...............`
(A)log(log x)+ c
(B) 1/2 (logx )2+c
(C) 2log x + c
(D) log x + c
Integrate the function in x log x.
Integrate the function in x log 2x.
Integrate the function in x (log x)2.
Integrate the function in e2x sin x.
Prove that:
`int sqrt(x^2 + a^2)dx = x/2 sqrt(x^2 + a^2) + a^2/2 log |x + sqrt(x^2 + a^2)| + c`
Evaluate the following:
`int sec^3x.dx`
Evaluate the following : `int e^(2x).cos 3x.dx`
Integrate the following functions w.r.t. x:
sin (log x)
Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`
Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`
Integrate the following w.r.t.x : log (log x)+(log x)–2
Evaluate the following.
`int "e"^"x" [(log "x")^2 + (2 log "x")/"x"]` dx
Evaluate the following.
`int [1/(log "x") - 1/(log "x")^2]` dx
`int ("x" + 1/"x")^3 "dx"` = ______
Evaluate `int 1/(x(x - 1)) "d"x`
Evaluate `int 1/(4x^2 - 1) "d"x`
`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.
Evaluate the following:
`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`
Evaluate the following:
`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.
The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1) dx` is
If u and v ore differentiable functions of x. then prove that:
`int uv dx = u intv dx - int [(du)/(d) intv dx]dx`
Hence evaluate `intlog x dx`
Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`
Find: `int e^x.sin2xdx`
Solve: `int sqrt(4x^2 + 5)dx`
`int e^x [(2 + sin 2x)/(1 + cos 2x)]dx` = ______.
`int1/sqrt(x^2 - a^2) dx` = ______
`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`
`int1/(x+sqrt(x)) dx` = ______
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((1 + sinx)/(1 + cosx))e^x dx`
Evaluate:
`int (sin(x - a))/(sin(x + a))dx`
Complete the following activity:
`int_0^2 dx/(4 + x - x^2) `
= `int_0^2 dx/(-x^2 + square + square)`
= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`
= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`
= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`
Evaluate the following.
`intx^3/sqrt(1+x^4)dx`
If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)
Evaluate.
`int(5x^2 - 6x + 3)/(2x - 3) dx`
