Advertisements
Advertisements
Question
Advertisements
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
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
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
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
Find: `int logx/(1 + log x)^2 dx`
