Advertisements
Advertisements
प्रश्न
`int ("x" + 1/"x")^3 "dx"` = ______
विकल्प
`1/4 ("x" + 1/"x")^4` + c
`"x"^4/4 + "3x"^2/2 + 3 log "x" - 1/"2x"^2 + "c"`
`"x"^4/4 + "3x"^2/2 + 3 log "x" + 1/"x"^2 + "c"`
`("x" - "x"^-1)^3` + c
Advertisements
उत्तर
`int ("x" + 1/"x")^3 "dx"` = `bbunderline("x"^4/4 + "3x"^2/2 + 3 log "x" - 1/"2x"^2 + "c")`
Explanation:
Let I = `int ("x" + 1/"x")^3 "dx"`
`int ("x"^3 + "3x" + 3/"x" + 1/"x"^3)` dx
`= "x"^4/4 + 3 "x"^2/2 + 3 log |"x"| - 1/"2x"^2` + c
APPEARS IN
संबंधित प्रश्न
Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`
Integrate the function in x log x.
Integrate the function in x log 2x.
Integrate the function in x (log x)2.
Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.
Evaluate the following : `int x^2.log x.dx`
Evaluate the following : `int x^2tan^-1x.dx`
Integrate the following functions w.r.t. x : `e^(2x).sin3x`
Integrate the following functions w.r.t. x : [2 + cot x – cosec2x]ex
Integrate the following functions w.r.t. x : `((1 + sin x)/(1 + cos x)).e^x`
Integrate the following w.r.t.x : log (x2 + 1)
Solve the following differential equation.
(x2 − yx2 ) dy + (y2 + xy2) dx = 0
Evaluate the following.
`int (log "x")/(1 + log "x")^2` dx
Evaluate:
∫ (log x)2 dx
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
`int ["cosec"(logx)][1 - cot(logx)] "d"x`
Choose the correct alternative:
`intx^(2)3^(x^3) "d"x` =
Choose the correct alternative:
`int ("d"x)/((x - 8)(x + 7))` =
`int (x^2 + x - 6)/((x - 2)(x - 1)) "d"x` = x + ______ + c
Find `int_0^1 x(tan^-1x) "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 - a^2)dx` = ______.
`inte^(xloga).e^x dx` is ______
Evaluate:
`int e^(ax)*cos(bx + c)dx`
Evaluate:
`int e^(logcosx)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.
`int x^3 e^(x^2) dx`
The value of `inta^x.e^x dx` equals
Evaluate the following.
`intx^3 e^(x^2)dx`
