Advertisements
Advertisements
प्रश्न
Evaluate the following.
`int e^x (1/x - 1/x^2)`dx
Advertisements
उत्तर
Let I = `int e^x (1/x - 1/x^2)`dx
Put f(x) = `1/x`
∴ f '(x) = `1/x`
∴ I = `int e^x [f(x) + f '(x)]` dx
`= e^x * f(x) + c`
∴ I = `e^x * 1/x + c`
∴ I = `(e^x)/x + c`
Notes
The answer in the textbook is incorrect.
APPEARS IN
संबंधित प्रश्न
Integrate the function in tan-1 x.
Find :
`∫(log x)^2 dx`
Evaluate the following:
`int x^2 sin 3x dx`
Evaluate the following : `int x^3.tan^-1x.dx`
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Integrate the following functions w.r.t. x : `e^(2x).sin3x`
Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`
Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`
Integrate the following functions w.r.t. x : `[x/(x + 1)^2].e^x`
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Choose the correct alternative from the following.
`int (("e"^"2x" + "e"^"-2x")/"e"^"x") "dx"` =
Evaluate:
∫ (log x)2 dx
`int(x + 1/x)^3 dx` = ______.
`int logx/(1 + logx)^2 "d"x`
`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.
`int log x * [log ("e"x)]^-2` dx = ?
`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.
Find `int_0^1 x(tan^-1x) "d"x`
The value of `int_0^(pi/2) log ((4 + 3 sin x)/(4 + 3 cos x)) dx` is
If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.
Evaluate :
`int(4x - 6)/(x^2 - 3x + 5)^(3/2) dx`
The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.
Evaluate:
`int e^(ax)*cos(bx + c)dx`
`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.
Evaluate the following.
`int x sqrt(1 + x^2) dx`
Evaluate the following.
`intx^3 e^(x^2)dx`
