Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ 1 + \log\ x\ = t . Then, \frac{1}{x}\ dx\ = dt\]
\[When\ x = 1, t = 1\ and\ x\ = 2, t = \left( 1 + \log 2 \right)\]
\[ \therefore I = \int_1^2 \frac{1}{x \left( 1 + \log x \right)^2} d x\]
\[ \Rightarrow I = \int_1^\left( 1 + \log 2 \right) \frac{1}{t^2} dt\]
\[ \Rightarrow I = \left[ \frac{- 1}{t} \right]_1^\left( 1 + \log 2 \right) \]
\[ \Rightarrow I = - \frac{1}{\left( 1 + \log 2 \right)} + 1\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2 + \log e}\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2e}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
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
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
