Advertisements
Advertisements
Question
Integrate the function `1/sqrt(7 - 6x - x^2)`
Advertisements
Solution
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
RELATED QUESTIONS
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Evaluate : ` int x^2/((x^2+4)(x^2+9))dx`
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `(3x^2)/(x^6 + 1)`
Integrate the function `1/sqrt(9 - 25x^2)`
Integrate the function `(3x)/(1+ 2x^4)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `1/(9x^2 + 6x + 5)`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
`int dx/(x^2 + 2x + 2)` equals:
`int dx/sqrt(9x - 4x^2)` equals:
Integrate the function:
`sqrt(4 - x^2)`
Integrate the function:
`sqrt(1- 4x^2)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
Find `int (2x)/(x^2 + 1)(x^2 + 2)^2 dx`
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
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)/sqrt(4x - x^2)`
Find: `int (dx)/(x^2 - 6x + 13)`
