Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{ x \text{log x } \text{log }\left( \text{log x }\right)} dx\]
बेरीज
उत्तर
` Note: "Here, we are considering log x as" log_e x . `
\[\text{Let I }= \int\frac{1}{x \log x \log\left( \log x \right)}dx\]
\[Putting \log\left( \log x \right) = t\]
\[ \Rightarrow \frac{1}{x\log x} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{x \log x}dx = dt\]
\[ \therefore I = \int\frac{dt}{t}\]
\[ = \log\left| t \right| + C\]
\[ = \log\left| \text{log}\left( \ logx \right) \right| + C \left[ \because t = \text{log}\left( \text{log x} \right) \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\frac{\left( 1 + x \right)^3}{\sqrt{x}} dx\]
\[\int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
\[\int \cos^7 x \text{ dx } \]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int \sin^7 x \text{ dx }\]
\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]
\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int x^2 \cos 2x\ \text{ dx }\]
\[\int x \cos^2 x\ dx\]
\[\int x \sin x \cos x\ dx\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int \log_{10} x\ dx\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int x\sqrt{x^4 + 1} \text{ dx}\]
\[\int\frac{1}{x^4 - 1} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]
\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int\frac{1}{7 + 5 \cos x} dx =\]
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
\[\int\frac{\sin x}{\cos 2x} \text{ dx }\]
\[\int x \sec^2 2x\ dx\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
\[\int\frac{x}{x^3 - 1} \text{ dx}\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
Find : \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\]