Advertisements
Advertisements
Question
Integrate the function `1/sqrt(x^2 +2x + 2)`
Solution
Let `I = int 1/ sqrt (x^2 + 2x + 2) dx`
`= dx/ sqrt ((x + 1)^2 + 1)`
∴ `I = log |(x + 1) + sqrt ((x + 1)^2 + 1)| + C` ` ....[∵ int dx/sqrt (a^2 + x^2) = log |x + sqrt (x^2 + a^2)| + C]`
`= log |(x + 1) + sqrt (x^2 + 2x + 2)| +C`
APPEARS IN
RELATED QUESTIONS
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `1/sqrt(1+4x^2)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Integrate the function `(5x - 2)/(1 + 2x + 3x^2)`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(x+2)/sqrt(x^2 + 2x + 3)`
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(1- 4x^2)`
Integrate the function:
`sqrt(x^2 + 4x + 6)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(x^2 + 3x)`
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
Find `int dx/(5 - 8x - x^2)`
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_(-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)`
