Advertisements
Advertisements
प्रश्न
Integrate the function `x^2/(1 - x^6)`
उत्तर
Let `I = x^2/(1 - x^6) dx`
`= int x^2/(1 - (x^3)^2) dx`
Put x3 = t
3x2 dx = dt ⇒ x2 dx = `1/3` dt
`therefore I = 1/3 int dt/(1 - t^2)`
`= 1/3 . 1/2 log abs ((1 + t)/(1 - t)) + C`
`= 1/6 log abs ((1 + t)/(1 - t)) + C`
`= 1/6 log abs ((1 + x^3)/(1 - x^3)) + C` `...[∵ int dx/(a^2 - x^2) = 1/(2a) log |(a + x) /(a - x)|+C]`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `1/sqrt(1+4x^2)`
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `(3x)/(1+ 2x^4)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `(sec^2 x)/sqrt(tan^2 x + 4)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(4x+ 1)/sqrt(2x^2 + x - 3)`
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(1-4x - x^2)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Integrate the function:
`sqrt(x^2 + 3x)`
Integrate the function:
`sqrt(1+ x^2/9)`
`int sqrt(1+ x^2) dx` is equal to ______.
Evaluate : `int_2^3 3^x dx`
Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
Find `int (dx)/sqrt(4x - x^2)`
Find: `int (dx)/(x^2 - 6x + 13)`
