Advertisements
Advertisements
Question
Evaluate the following.
∫ x log x dx
Solution
Let I = ∫ x log x dx
`= log "x" int "x" "dx" - int["d"/"dx" (log "x") int "x dx"] "dx"`
`= log "x" * "x"^2/2 - int [1/"x" xx "x"^2/2]` dx
`= "x"^2/2 log "x" - 1/2 int "x dx"`
`= "x"^2/2 log "x" - 1/2 * "x"^2/2 + "c"`
∴ I = `"x"^2/2 log "x" - "x"^2/4 + "c"`
Notes
The answer in the textbook is incorrect.
APPEARS IN
RELATED QUESTIONS
If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:
(A) 0
(B) π
(C) π/2
(D) π/4
Integrate the function in tan-1 x.
Evaluate the following : `int x^3.tan^-1x.dx`
Evaluate the following : `int e^(2x).cos 3x.dx`
Choose the correct options from the given alternatives :
`int [sin (log x) + cos (log x)]*dx` =
Evaluate: `int "e"^"x"/(4"e"^"2x" -1)` dx
`int ("d"x)/(x - x^2)` = ______
`int logx/(1 + logx)^2 "d"x`
`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?
∫ log x · (log x + 2) dx = ?
Evaluate the following:
`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`
The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1) dx` is
Find: `int e^x.sin2xdx`
Solve: `int sqrt(4x^2 + 5)dx`
The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.
Solution of the equation `xdy/dx=y log y` is ______
`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 the following.
`intx^3e^(x^2) dx`
