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: `int1/(xlogxlog(logx))dx`
Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`
Evaluate `int_(-1)^2|x^3-x|dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate the integral by using substitution.
`int_0^1 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_0^1 sin^(-1) ((2x)/(1+ x^2)) dx`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
Evaluate of the following integral:
(i) \[\int x^4 dx\]
Evaluate of the following integral:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" d"x"`.
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?
`int_0^1 x^2e^x dx` = ______.
After finding the antiderivative in Method: Resubstitution, what is the next step?
After changing the limits to the new variable, how is the definite integral completed?
After using \[t=\tan^{-1}x\] in \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which integral results?
