Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{Here, we are considering }\text{log x as} \log_e x . \]
\[\text{Let I} = \int\frac{1}{x \log x}dx\]
\[\text{Putting }\log x = t\]
\[ \Rightarrow \frac{1}{x} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{x}dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{log} \left| \text{log x} \right| + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^3dx/(9+x^2)`
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
Evaluate the following integrals:
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate: \[\int\frac{x^3 - x^2 + x - 1}{x - 1} \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
