Advertisements
Advertisements
प्रश्न
Evaluate each of the following integral:
Advertisements
उत्तर
\[\int_e^{e^2} \frac{1}{x\log x}dx\]
\[ = \int_e^{e^2} \frac{\frac{1}{x}}{\log x}dx\]
\[ = \left.\log\left( \log x \right)\right|_e^{e^2} ...............\left[ \int\frac{f'\left( x \right)}{f\left( x \right)}dx = \log f\left( x \right) + C \right]\]
\[ = \log\left( \log e^2 \right) - \log\left( \log e \right)\]
\[ = \log\left( 2\log e \right) - \log\left( \log e \right) \]
\[ = \log2 - \log1 ................\left( \log e = 1 \right)\]
\[ = \log2 - 0\]
\[ = \log2\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
