Advertisements
Advertisements
प्रश्न
Integrate the function `1/sqrt(7 - 6x - x^2)`
Advertisements
उत्तर
Let `I = int 1/sqrt(7 - 6x - x^2) dx`
`= dx/sqrt(7 - (x^2 + 6x))`
`= int dx/sqrt(7 - (x^2 + 6x + 9) + 9)`
`= int dx/sqrt(16 - (x + 3)^2)`
`= int dx/sqrt(4^2 - (x + 3)^2)`
`= sin^-1 ((x + 3)/4) + C` `...[because 1/sqrt(a^2 - x^2) dx = sin^-1 x/a + C]`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Integrate the function `(3x^2)/(x^6 + 1)`
Integrate the function `1/sqrt(1+4x^2)`
Integrate the function `1/sqrt(9 - 25x^2)`
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/sqrt((x -1)(x - 2))`
Integrate the function `1/sqrt(8+3x - x^2)`
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 `(6x + 7)/sqrt((x - 5)(x - 4))`
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(4 - 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)`
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
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 ______.
