Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^1 \log\left( \frac{1}{x} - 1 \right) d x ...............(1)\]
\[ = \int_0^1 \log\left( \frac{1}{1 - x} - 1 \right) d x ...............\left[\text{Using }\int_0^a f(x) dx = \int_0^a f(a - x) dx \right]\]
\[ I = \int_0^1 \log\left( \frac{x}{1 - x} \right) dx ...............(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^1 \log\left( \frac{1 - x}{x} \right) + \log\left( \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log\left( \frac{1 - x}{x} \times \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log1 dx \]
\[ = 0\]
\[Hence\ I = 0\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
`Γ(3/2)`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
