Advertisements
Advertisements
Question
Evaluate `int 1/(x(x - 1)) "d"x`
Advertisements
Solution
Let I = `int 1/(x(x - 1)) "d"x`
= `int(x - x + 1)/(x(x - 1)) "d"x`
= `int(x - (x - 1))/(x(x - 1)) "d"x`
= `int(1/(x - 1) - 1/x) "d"x`
= `int 1/(x - 1) "d"x - int 1/x "d"x`
= `log |x - 1| - log |x| + "c"`
∴ I = `log |(x - 1)/x| + "c"`
APPEARS IN
RELATED QUESTIONS
Integrate the function in x sec2 x.
Integrate the function in tan-1 x.
Integrate the function in `((x- 3)e^x)/(x - 1)^3`.
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 x.sin^-1 x.dx`
Evaluate the following : `int sin θ.log (cos θ).dθ`
Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`
Integrate the following functions w.r.t. x : `(x + 1) sqrt(2x^2 + 3)`
Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`
Integrate the following functions w.r.t. x : `e^(sin^-1x)*[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`
Choose the correct options from the given alternatives :
`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =
Choose the correct options from the given alternatives :
`int [sin (log x) + cos (log x)]*dx` =
Integrate the following with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Integrate the following w.r.t.x : log (log x)+(log x)–2
Integrate the following w.r.t.x : log (x2 + 1)
Integrate the following w.r.t.x : e2x sin x cos x
Evaluate the following.
`int e^x (1/x - 1/x^2)`dx
Evaluate the following.
`int "e"^"x" "x"/("x + 1")^2` dx
Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx
`int (sinx)/(1 + sin x) "d"x`
State whether the following statement is True or False:
If `int((x - 1)"d"x)/((x + 1)(x - 2))` = A log|x + 1| + B log|x – 2|, then A + B = 1
The value of `int_0^(pi/2) log ((4 + 3 sin x)/(4 + 3 cos x)) dx` 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
`inte^(xloga).e^x dx` is ______
Evaluate:
`int e^(ax)*cos(bx + c)dx`
Evaluate `int (1 + x + x^2/(2!))dx`
Evaluate:
`int x^2 cos x dx`
Which differentiation identity verifies the special integral \[\int e^x\left[f(x)+f'(x)\right]dx=e^xf(x)+C?\]
