Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_0^1 \frac{x}{x + 1} d x . Then, \]
\[I = \int_0^1 1 - \frac{1}{x + 1} d x\]
\[ \Rightarrow I = \left[ x - \log \left( x + 1 \right) \right]_0^1 \]
\[ \Rightarrow I = 1 - \log 2 - (0 - \log 1)\]
\[ \Rightarrow I = \log e - \log 2\]
\[ \Rightarrow I = \log \frac{e}{2}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
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
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Choose the correct alternative:
Γ(n) is
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
