Advertisements
Advertisements
प्रश्न
उत्तर
\[\int \log_{10} x \text{ dx }\]
\[ = \int\frac{\log x}{\log 10}dx\]
\[ = \frac{1}{\log 10}\int 1_{} \cdot \text{ log x dx }\]
` " Taking log x as the first function and 1 as the second function " `
\[ = \frac{1}{\log 10}\left[ \log x \int\text{ 1 dx} - \int\left\{ \frac{d}{dx}\left( \log x \right)\int\text{ 1 dx }\right\}dx \right]\]
\[ = \frac{1}{\log 10}\left[ \log x \cdot x - \int\frac{1}{x} \cdot \text{ x dx } \right]\]
\[ = \frac{1}{\log 10}\left[ x \log x - x \right] + C\]
\[ = \frac{1}{\log 10}\left[ x\left( \log x - 1 \right) \right] + C\]
APPEARS IN
संबंधित प्रश्न
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int {cosec}^4 2x\ dx\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
