Advertisements
Advertisements
प्रश्न
Evaluate:
Advertisements
उत्तर
\[\int\frac{e\log\sqrt{x}}{x}dx\]
\[ = \int\frac{\sqrt{x}}{x}dx\]
\[ = \int\frac{1}{\sqrt{x}} dx\]
\[ = \int x^{- \frac{1}{2}} dx\]
\[ = \left[ \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = 2\sqrt{x} + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate : `int1/(3+5cosx)dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate: `intsinsqrtx/sqrtxdx`
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_1^2 (1/x- 1/(2x^2))e^(2x) dx`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
`int 1/(1 + cos x)` dx = _____
A) `tan(x/2) + c`
B) `2 tan (x/2) + c`
C) -`cot (x/2) + c`
D) -2 `cot (x/2)` + c
Evaluate of the following integral:
Evaluate:
Evaluate :
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 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 the following integral:
Find : \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] .
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_ e^x ((2+sin2x))/cos^2 x dx`
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
Find: `int (dx)/sqrt(3 - 2x - x^2)`
`int_0^1 x^2e^x dx` = ______.
Evaluate: `int x/(x^2 + 1)"d"x`
