Advertisements
Advertisements
प्रश्न
Integrate the function `1/sqrt((2-x)^2 + 1)`
Advertisements
उत्तर
Let `I = int 1/sqrt((2 - x)^2 + 1) dx`
Put 2 - x = t
- dx = dt ⇒ dx = - dt
`therefore I = - int dt/sqrt(t^2 + 1) dt`
`= - log [t + sqrt(t^2 + 1)] + C`
`= - log [(2 - x) + sqrt((2 - x)^2 + 1)] + C`
`= log |1/ ((2 - x) + sqrt (x^2 - 4x + 5))| + 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 `(3x^2)/(x^6 + 1)`
Integrate the function `(3x)/(1+ 2x^4)`
Integrate the function `x^2/(1 - x^6)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
`int dx/(x^2 + 2x + 2)` equals:
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(x^2 + 4x + 6)`
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)`
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`
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"`
If θ f(x) = `int_0^x t sin t dt` then `f^1(x)` is
Find `int (dx)/sqrt(4x - x^2)`
Find: `int (dx)/(x^2 - 6x + 13)`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
