Advertisements
Advertisements
प्रश्न
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
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
