Advertisements
Advertisements
प्रश्न
`intx^2 e^(x^3) dx` equals:
विकल्प
`1/3 e^(x^3) + C`
`1/3 e^(x^2) + C`
`1/2 e^(x^3) + C`
`1/2 e^(x^2) + C`
उत्तर
`1/3 e^(x^3) + C`
स्पष्टीकरण:
`int x^2 e^(x^3)` dx
Putting x3 = t, 3x2 dx = dt
`= 1/3 int (3x^2)e^(x^3)` dx
`= 1/3 int e^t dt = 1/3 e^t + C`
`= 1/3 e^(x^3) + C`
APPEARS IN
संबंधित प्रश्न
Integrate the function in x sin x.
Integrate the function in x tan-1 x.
Integrate the function in x sec2 x.
Integrate the function in ex (sinx + cosx).
Find :
`∫(log x)^2 dx`
Evaluate the following : `int x tan^-1 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 : `e^(2x).sin3x`
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 functions w.r.t. x : `xsqrt(5 - 4x - x^2)`
Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`
Integrate the following functions w.r.t. x : [2 + cot x – cosec2x]ex
Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`
Choose the correct options from the given alternatives :
`int tan(sin^-1 x)*dx` =
Integrate the following w.r.t.x : `(1)/(xsin^2(logx)`
Evaluate: `int "dx"/(5 - 16"x"^2)`
Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx
`int 1/(4x + 5x^(-11)) "d"x`
`int ["cosec"(logx)][1 - cot(logx)] "d"x`
`int 1/x "d"x` = ______ + c
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
`int cot "x".log [log (sin "x")] "dx"` = ____________.
Evaluate the following:
`int_0^1 x log(1 + 2x) "d"x`
The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1) dx` is
`int 1/sqrt(x^2 - 9) dx` = ______.
Find `int e^(cot^-1x) ((1 - x + x^2)/(1 + x^2))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
The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.
`int(xe^x)/((1+x)^2) dx` = ______
Evaluate:
`int e^(logcosx)dx`
Evaluate:
`int (logx)^2 dx`
Evaluate `int tan^-1x dx`
If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv dx - int(d/dx u)(intv dx)dx`. Hence evaluate: `intx cos x dx`
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`
