Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} d x . Then, \]
\[I = \int_e^{e^2} 1 \frac{1}{\log x} dx - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[\text{Integrating by parts}\]
\[ \Rightarrow I = \left\{ \left[ \frac{x}{\log x} \right]_e^{e^2} - \int_e^{e^2} \frac{- 1}{x \left( \log x \right)^2} x d x \right\} - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[ \Rightarrow I = \left[ \frac{x}{\log x} \right]_e^{e^2} + \int_e^{e^2} \frac{1}{\left( \log x \right)^2} d x - \int_e^{e^2} \frac{1}{\left( \log x \right)^2} dx\]
\[ \Rightarrow I = \left[ \frac{x}{\log x} \right]_e^{e^2} + 0\]
\[ \Rightarrow I = \frac{e^2}{\log e^2} - \frac{e}{\log e}\]
\[ \Rightarrow I = \frac{e^2}{2 \log e} - \frac{e}{\log e}\]
\[ \Rightarrow I = \frac{e^2}{2} - e\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
If f is an integrable function, show that
Solve each of the following integral:
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
