Advertisements
Advertisements
प्रश्न
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
Advertisements
उत्तर
`int_-1^1 dx/(x^2 + 2x + 5)`
`= int_-1^1 dx/(x^2 + 2x + 4 + 1)`
`int_-1^1 dx/((x + 2)^2 + (1)^2)`
`= 1/2 [tan^-1 ((x + 1)/2)]_-1^1`
`= 1/2 [tan^-1 (1) - tan^-1 0]`
`= 1/2 [pi/4 - 0]`
`= pi/8`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`
Evaluate `∫_0^(3/2)|x cosπx|dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate the integral by using substitution.
`int_0^(pi/2) sqrt(sin phi) cos^5 phidphi`
Evaluate the integral by using substitution.
`int_0^2 xsqrt(x+2)` (Put x + 2 = `t^2`)
Evaluate the integral by using substitution.
`int_0^(pi/2) (sin x)/(1+ cos^2 x) dx`
Evaluate the integral by using substitution.
`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate the following integral:
\[\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx\]
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
Find: `int_ (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`int_0^1 x(1 - x)^5 "dx" =` ______.
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
`int_0^(pi4) sec^4x "d"x` = ______.
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
Find: `int (dx)/sqrt(3 - 2x - x^2)`
Evaluate: `int x/(x^2 + 1)"d"x`
