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:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
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 using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
If n > 0, then Γ(n) is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
