Advertisements
Advertisements
प्रश्न
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
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
Evaluate :
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
