Advertisements
Advertisements
प्रश्न
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
विकल्प
1
0
−1
π/4
Advertisements
उत्तर
0
\[Let\, I = \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} d x ................(1)\]
\[ = \int_0^1 \tan^{- 1} \frac{2\left( 1 - x \right) - 1}{1 + \left( 1 - x \right) - \left( 1 - x \right)^2} d x\]
\[ = \int_0^1 \tan^{- 1} \frac{1 - 2x}{2 - x - 1 - x^2 + 2x} dx\]
\[ = \int_0^1 \tan^{- 1} \frac{1 - 2x}{1 + x - x^2} dx\]
\[ = - \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx .................(2)\]
\[\text{Adding (i) and (ii)}\]
\[2I = \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx - \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx\]
\[ = 0\]
\[Hence\, I = 0\]
APPEARS IN
संबंधित प्रश्न
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
Evaluate :
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
Γ(1) is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
