Advertisements
Advertisements
प्रश्न
If u and v are two functions of x then prove that
`intuvdx=uintvdx-int[du/dxintvdx]dx`
Hence evaluate, `int xe^xdx`
Advertisements
उत्तर
Let ` int vdx=w.....(1)`
`then " " (dw)/dx=v.....(2)`
`Now d/dx(u,w)=u.d/dx(w)+wd/dx(u)`
`=u.v+w(du)/dx......."from"(2)`
By definition of integration.
`u.w=int[u.v+w(du)/dx]dx`
`=intu.vdx+intw.(du)/dx dx`
`int u.v dx=u.w-int w (du)/dx dx`
`=u int v dx-int [(du)/dxintv.dx]dx`
[next section only required for question 2]
Hence, `int xe^xdx = x.inte^xdx-int[d/dx x.inte^xdx]dx`
`=xe^x-int1xxe^xdx`
`=xe^x-e^x+c`
APPEARS IN
संबंधित प्रश्न
Integrate the function in x2 log x.
Integrate the function in x sin−1 x.
Integrate the function in x tan-1 x.
Integrate the function in x cos-1 x.
Integrate the function in `(xe^x)/(1+x)^2`.
Evaluate the following:
`int x tan^-1 x . dx`
Evaluate the following: `int x.sin^-1 x.dx`
Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`
Integrate the following functions w.r.t. x : `e^(2x).sin3x`
Integrate the following functions w.r.t.x:
`e^(5x).[(5x.logx + 1)/x]`
Choose the correct options from the given alternatives :
`int (log (3x))/(xlog (9x))*dx` =
Choose the correct options from the given alternatives :
`int tan(sin^-1 x)*dx` =
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`
Integrate the following w.r.t.x : log (x2 + 1)
Evaluate the following.
`int [1/(log "x") - 1/(log "x")^2]` dx
Evaluate the following.
`int (log "x")/(1 + log "x")^2` dx
`int ("x" + 1/"x")^3 "dx"` = ______
Evaluate: `int "dx"/("9x"^2 - 25)`
`int 1/(4x + 5x^(-11)) "d"x`
Evaluate `int 1/(x(x - 1)) "d"x`
Evaluate `int (2x + 1)/((x + 1)(x - 2)) "d"x`
`int logx/(1 + logx)^2 "d"x`
`int 1/sqrt(x^2 - 8x - 20) "d"x`
Evaluate the following:
`int_0^pi x log sin x "d"x`
Find: `int e^x.sin2xdx`
Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`
`int 1/sqrt(x^2 - a^2)dx` = ______.
Evaluate `int(3x-2)/((x+1)^2(x+3)) 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 (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`
Evaluate the following.
`intx^3 e^(x^2) dx`
Evaluate the following.
`intx^3/sqrt(1+x^4)`dx
The value of `inta^x.e^x dx` equals
Evaluate the following.
`intx^3 e^(x^2)dx`
Evaluate the following.
`intx^3/(sqrt(1 + x^4))dx`
