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
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
