Advertisements
Advertisements
Question
Integrate the function `1/sqrt((2-x)^2 + 1)`
Advertisements
Solution
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
RELATED QUESTIONS
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`
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 `(3x)/(1+ 2x^4)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `1/sqrt(7 - 6x - x^2)`
Integrate the function `1/sqrt((x -1)(x - 2))`
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))`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
Integrate the function:
`sqrt(4 - x^2)`
Integrate the function:
`sqrt(x^2 + 3x)`
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
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
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
Find `int (dx)/sqrt(4x - x^2)`
Find: `int (dx)/(x^2 - 6x + 13)`
