Advertisements
Advertisements
Question
Solution
\[Let\ 1 + \log\ x\ = t . Then, \frac{1}{x}\ dx\ = dt\]
\[When\ x = 1, t = 1\ and\ x\ = 2, t = \left( 1 + \log 2 \right)\]
\[ \therefore I = \int_1^2 \frac{1}{x \left( 1 + \log x \right)^2} d x\]
\[ \Rightarrow I = \int_1^\left( 1 + \log 2 \right) \frac{1}{t^2} dt\]
\[ \Rightarrow I = \left[ \frac{- 1}{t} \right]_1^\left( 1 + \log 2 \right) \]
\[ \Rightarrow I = - \frac{1}{\left( 1 + \log 2 \right)} + 1\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2 + \log e}\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2e}\]
APPEARS IN
RELATED QUESTIONS
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
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 \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
