Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^1 \frac{1 - x}{1 + x} d\ x\ . Then, \]
\[I = \int_0^1 \left( \frac{1}{1 + x} - \frac{1 + x - 1}{1 + x} \right) d x\]
\[I = \int_0^1 \left( \frac{1}{1 + x} - 1 + \frac{x}{1 + x} \right) d x\]
\[ \Rightarrow I = \left[ \log \left( 1 + x \right) - x + \log \left( 1 + x \right) \right]_0^1 \]
\[ \Rightarrow I = \left( \log 2 - 1 + \log 2 \right) - \left( \log 1 - 0 + \log 1 \right)\]
\[ = 2 \log 2 - 1\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
If f is an integrable function, show that
Evaluate each of the following integral:
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
`int x^3/(x + 1)` is equal to ______.
