Advertisements
Advertisements
प्रश्न
Integrate the function `x^2/(1 - x^6)`
Advertisements
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 x^2/((x^2+4)(x^2+9))dx`
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `(3x)/(1+ 2x^4)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(1- 4x^2)`
Integrate the function:
`sqrt(x^2 + 4x + 6)`
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Integrate the function:
`sqrt(x^2 + 3x)`
`int sqrt(1+ x^2) dx` is equal to ______.
Find `int dx/(5 - 8x - x^2)`
Evaluate : `int_2^3 3^x dx`
Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
Find: `int (dx)/(x^2 - 6x + 13)`
