Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\, I = \int\frac{1 - x^2}{x^4 + x^2 + 1} dx\]
\[ = - \int\frac{x^2 - 1}{x^4 + x^2 + 1} dx\]
\[ = - \int\frac{1 - \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} dx\]
\[ = - \int\frac{1 - \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2} - 1} dx\]
\[ = - \int\frac{1 - \frac{1}{x^2}}{\left( x + \frac{1}{x} \right)^2 - 1} dx\]
\[Let, x + \frac{1}{x} = t\]
\[ \Rightarrow \left( 1 - \frac{1}{x^2} \right)dx = dt\]
\[\text{Then integral becomes}, \]
\[I = - \int\frac{1}{t^2 - 1} dt\]
\[ = - \frac{1}{2}\log\left| \frac{t - 1}{t + 1} \right|\]
\[ = \frac{1}{2}\log\left| \frac{t + 1}{t - 1} \right|\]
\[ = \frac{1}{2}\log\left| \frac{x + \frac{1}{x} + 1}{x + \frac{1}{x} - 1} \right|\]
\[ = \frac{1}{2}\log\left| \frac{x^2 + x + 1}{x^2 - x + 1} \right|\]
\[i . e . , \int\frac{1 - x^2}{x^4 + x^2 + 1} dx = \frac{1}{2}\log\left| \frac{x^2 + x + 1}{x^2 - x + 1} \right|\]
\[ \Rightarrow \int_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx = \left[ \frac{1}{2}\log\left| \frac{x^2 + x + 1}{x^2 - x + 1} \right| \right]_0^1 \]
\[ = \frac{1}{2}\log 3\]
APPEARS IN
संबंधित प्रश्न
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
Γ(n) is
Which integral has lower and upper limits?
