Advertisements
Advertisements
प्रश्न
Integrate the function `1/sqrt(1+4x^2)`
Advertisements
उत्तर
Let `I = int 1/sqrt(1 + 4x^2) dx`
`= 1/2 int dx/sqrt(1/4 + x)`
`= 1/2 int dx/sqrt((1/2)^2 + x^2)`
`= 1/2 log abs (x sqrt(1/4 +x^2)) C_1`
.....`[because int dx/sqrt(x^2 + a^2) = log abs (x + sqrt(x^2 + a^2)) + C]`
`= 1/2 log abs ((2x + sqrt(1 + 4x^2))/2) + C_1`
`= 1/2 log abs (2x + sqrt(1 + 4x^2)) - 1/2 log 2 + C_1`
`= 1/2 log |2x + sqrt 1 + 4x^2| + C` .... [`C = -1/2 log 2 + C_1`]
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 `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt(7 - 6x - x^2)`
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(x^2 + 2x + 3)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(x^2 + 4x + 6)`
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 ______.
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
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)/(x^2 - 6x + 13)`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
