Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} d x . Then, \]
\[I = \int_e^{e^2} 1 \frac{1}{\log x} dx - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[\text{Integrating by parts}\]
\[ \Rightarrow I = \left\{ \left[ \frac{x}{\log x} \right]_e^{e^2} - \int_e^{e^2} \frac{- 1}{x \left( \log x \right)^2} x d x \right\} - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[ \Rightarrow I = \left[ \frac{x}{\log x} \right]_e^{e^2} + \int_e^{e^2} \frac{1}{\left( \log x \right)^2} d x - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[ \Rightarrow I = \left[ \frac{x}{\log x} \right]_e^{e^2} + 0\]
\[ \Rightarrow I = \frac{e^2}{\log e^2} - \frac{e}{\log e}\]
\[ \Rightarrow I = \frac{e^2}{2 \log e} - \frac{e}{\log e}\]
\[ \Rightarrow I = \frac{e^2}{2} - e\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
Find: `int logx/(1 + log x)^2 dx`
