Advertisements
Advertisements
Question
\[\int\frac{\log \left( \log x \right)}{x} dx\]
Sum
Solution
\[\int\frac{\log \left( \log x \right)}{x}dx\]
` "Taking log log x as the first function and "1/x "as the second function" . `
\[ = \text{ log }\log x\int\frac{1}{x}dx - \int\left\{ \frac{d}{dx} \text{ log }\left( \log x \right)\int\frac{1}{x}dx \right\}dx\]
\[ = \log x . \text{ log }\left( \log x \right) - \int\frac{1}{x \log x}\left( \log x \right)dx\]
\[ = \log x . \text{ log }\left( \log x \right) - \int\frac{1}{x}dx\]
\[ = \log x . \text{ log }\left( \log x \right) - \log x + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
\[\int \sin^2 \frac{x}{2} dx\]
` ∫ sin 4x cos 7x dx `
\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\]
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
\[\int x^3 \sin x^4 dx\]
\[\ ∫ x \text{ e}^{x^2} dx\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
\[\int \cot^6 x \text{ dx }\]
\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]
\[\int\frac{1}{\sqrt{a^2 - b^2 x^2}} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{1}{\sqrt{5 - 4x - 2 x^2}} dx\]
\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]
\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{1}{1 + 3 \sin^2 x} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int \left( \log x \right)^2 \cdot x\ dx\]
\[\int\cos\sqrt{x}\ dx\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
\[\int\frac{1}{1 - \cos x - \sin x} dx =\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int \log_{10} x\ dx\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
