Advertisements
Advertisements
प्रश्न
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
पर्याय
- \[\frac{\pi}{60}\]
- \[\frac{\pi}{20}\]
- \[\frac{\pi}{40}\]
- \[\frac{\pi}{80}\]
उत्तर
\[\frac{\pi}{60}\]
\[ \int_0^\infty \frac{1}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)} d x\]
\[ = \frac{1}{5} \int_0^\infty \frac{1}{\left( x^2 + 4 \right)} - \frac{1}{\left( x^2 + 9 \right)}dx\]
\[ = \frac{1}{5} \left[ \frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{3} \tan^{- 1} \frac{x}{3} \right]_0^\infty \]
\[ = \frac{1}{5}\left[ \frac{1}{2} \times \frac{\pi}{2} - \frac{1}{3} \times \frac{\pi}{2} \right]\]
\[ = \frac{1}{5} \times \frac{\pi}{12}\]
\[ = \frac{\pi}{60}\]
APPEARS IN
संबंधित प्रश्न
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
Evaluate :
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`