Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_1^3 \frac{\log x}{\left( 1 + x \right)^2} d\ x\ . Then, \]
\[I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 - \int_1^3 \frac{1}{x}\left( \frac{- 1}{x + 1} \right) d x\]
\[ \Rightarrow I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 + \int_1^3 \frac{1}{x\left( x + 1 \right)} dx\]
\[ \Rightarrow I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 + \int_1^3 \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx\]
\[ \Rightarrow I = \left[ \frac{- 1}{1 + x} \log x \right]_1^3 + \left[ \log x - \log \left( x + 1 \right) \right]_1^3 \]
\[ \Rightarrow I = \frac{- 1}{4} \log 3 + \log 3 - \log 4 + \log 2\]
\[ \Rightarrow I = \frac{3}{4} \log 3 - \log 2\]
APPEARS IN
संबंधित प्रश्न
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f(x) is a continuous function defined on [−a, a], then prove that
Evaluate each of the following integral:
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) 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
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Choose the correct alternative:
Γ(n) is
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
