Advertisements
Advertisements
प्रश्न
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
उत्तर
\[\text{ Let I } = \int\frac{\log x}{x} dx\]
\[\text{ and }\text{ let} \log x = t\]
\[ \Rightarrow \frac{1}{x} dx = dt\]
\[ \therefore I = \int t \cdot dt\]
\[ = \frac{t^2}{2} + C\]
\[ = \frac{\left( \log x \right)^2}{2} + C \left( \because t = \log x \right)\]
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate:
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
