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
- \[\frac{\pi}{60}\]
- \[\frac{\pi}{20}\]
- \[\frac{\pi}{40}\]
- \[\frac{\pi}{80}\]
Advertisements
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
Evaluate the following definite integrals:
Evaluate each of the following integral:
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
`int x^3/(x + 1)` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
