Advertisements
Advertisements
प्रश्न
Integrate the function:
`sqrt(x^2 + 3x)`
Advertisements
उत्तर
Let `I = int sqrt(x^2 + 3x)` dx
`= int sqrt((x^2 + 3x + 9/4) - 9/4)` dx
`= int sqrt((x + 3/2)^2 - (3/2)^2)` dx
`= ((x + 3/2))/2 sqrt((x + 3/2)^2 - 9/4) - 9/8 log abs ((x + 3/2) + sqrt((x + 3/2)^2 - 9/4)) + C` `....[∵ int sqrt (x^2 - a^2) dx = x/2 sqrt (x^2 - a^2) - a^2/2 log |x + sqrt (x^2 - a^2)| + C]`
`= (2x + 3)/4 sqrt (x^2 + 3x) - 9/8 log abs (x + 3/2 + sqrt(x^2 + 3x)) + C`
APPEARS IN
संबंधित प्रश्न
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`
Integrate the function `(3x^2)/(x^6 + 1)`
Integrate the function `1/sqrt((2-x)^2 + 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 `x^2/sqrt(x^6 + a^6)`
Integrate the function `(sec^2 x)/sqrt(tan^2 x + 4)`
Integrate the function `1/(9x^2 + 6x + 5)`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(1-4x - x^2)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Find `int dx/(5 - 8x - x^2)`
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_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
Find `int (dx)/sqrt(4x - x^2)`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
