Advertisements
Advertisements
प्रश्न
`int(x + 1/x)^3 dx` = ______.
विकल्प
`1/4(x + 1/x)^4 + c`
`x^4/4 + (3x^2)/2 + 3log x - 1/(2x^2) + c`
`x^4/4 + (3x^2)/2 + 3log x + 1/x^2 + c`
`(x - x^(-1))^3 + c`
उत्तर
`int(x + 1/x)^3 dx` = `bb(underline(x^4/4 + (3x^2)/2 + 3log x - 1/(2x^2) + c))`.
Explanation:
`(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3`
∴ `int(x + 1/x)^3dx = int(x^3 + 3x + 3/x + 1/x^3)dx`
= `x^4/4 + (3x^2)/2 + 3logx - 1/(2x^2) + c`
APPEARS IN
संबंधित प्रश्न
Integrate the function in x sin 3x.
Integrate the function in x tan-1 x.
Integrate the function in x cos-1 x.
Integrate the function in tan-1 x.
Integrate the function in `(xe^x)/(1+x)^2`.
Evaluate the following : `int x^3.logx.dx`
Evaluate the following : `int cos sqrt(x).dx`
Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`
Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`
Integrate the following functions w.r.t. x : `e^(sin^-1x)*[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`
Integrate the following with respect to the respective variable : cos 3x cos 2x cos x
Integrate the following w.r.t.x : sec4x cosec2x
Evaluate: `int "dx"/(3 - 2"x" - "x"^2)`
`int (sinx)/(1 + sin x) "d"x`
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
Choose the correct alternative:
`intx^(2)3^(x^3) "d"x` =
`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?
`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.
Evaluate the following:
`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`
`int e^x [(2 + sin 2x)/(1 + cos 2x)]dx` = ______.
Evaluate :
`int(4x - 6)/(x^2 - 3x + 5)^(3/2) dx`
`int(1-x)^-2 dx` = ______
`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`
Evaluate:
`int((1 + sinx)/(1 + cosx))e^x dx`
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:
`int x^2 cos x dx`
Evaluate the following.
`int x^3 e^(x^2) dx`
The value of `inta^x.e^x dx` equals
