Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 \frac{1}{2 x^2 + x + 1} d\ x . Then, \]
\[I = \frac{1}{2} \int_0^1 \frac{1}{x^2 + \frac{x}{2} + \frac{1}{2}} d x\]
\[I = \frac{1}{2} \int_0^1 \frac{1}{\left( x^2 + \frac{x}{2} + \frac{1}{16} \right) - \frac{1}{16} + \frac{1}{2}} d\ x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^1 \frac{1}{\left( x + \frac{1}{4} \right)^2 + \frac{7}{16}} dx\]
\[ \Rightarrow I = \frac{1}{2} \times \frac{4}{\sqrt{7}} \left[ \tan^{- 1} \left( \frac{x + \frac{1}{4}}{\frac{\sqrt{7}}{4}} \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{2}{\sqrt{7}}\left( \tan^{- 1} \frac{5}{\sqrt{7}} - \tan^{- 1} \frac{1}{\sqrt{7}} \right)\]
APPEARS IN
संबंधित प्रश्न
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
