Advertisements
Advertisements
प्रश्न
Evaluate of the following integral:
Advertisements
उत्तर
\[ = \int\frac{dx}{x^{2/3}}\]
\[ = \int x^{- 2/3} dx\]
\[ = \frac{x^{- \frac{2}{3} + 1}}{- \frac{2}{3} + 1} + C\]
\[ = 3 x^\frac{1}{3} + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate :`int_0^(pi/2)1/(1+cosx)dx`
Evaluate : `int1/(3+5cosx)dx`
Evaluate `int_(-1)^2|x^3-x|dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate the integral by using substitution.
`int_0^2 dx/(x + 4 - x^2)`
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
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:
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 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_-1^2 (|"x"|)/"x"d"x"`.
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`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` = ______.
Evaluate the following:
`int "dt"/sqrt(3"t" - 2"t"^2)`
Find: `int (dx)/sqrt(3 - 2x - x^2)`
`int_0^1 x^2e^x dx` = ______.
