Advertisements
Advertisements
प्रश्न
\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]
बेरीज
Advertisements
उत्तर
\[\text{ Let I }= \int\sqrt{\frac{16 + \left( \log x \right)^2}{x}}\text{ dx}\]
\[\text{ Putting log x }= t\]
\[ \Rightarrow \frac{1}{x} \text{ dx}= dt\]
\[ \therefore I = \int\sqrt{16 + t^2}dt\]
\[ = \int\sqrt{4^2 + t^2}dt\]
\[ = \frac{t}{2} \sqrt{4^2 + t^2} + \frac{4^2}{2} \text{ log} \left| t + \sqrt{4^2 + t^2} \right| + C\]
\[ = \frac{\log x}{2} \sqrt{16 + \left( \log x \right)^2} + 8 \text{ log }\left| \log x + \sqrt{16 + \left( \log x \right)^2} \right| + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]
\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]
`∫ cos ^4 2x dx `
` ∫ cos 3x cos 4x` dx
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]
` ∫ sec^6 x tan x dx `
` ∫ \sqrt{tan x} sec^4 x dx `
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int \sin^5 x \text{ dx }\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int\frac{e^x}{1 + e^{2x}} dx\]
\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
` ∫ \sqrt{"cosec x"- 1} dx `
\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]
\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int x^3 \cos x^2 dx\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{2x + 1}{\left( x + 1 \right) \left( x - 2 \right)} dx\]
\[\int\frac{\sin 2x}{\left( 1 + \sin x \right) \left( 2 + \sin x \right)} dx\]
\[\int\sqrt{\cot \text{θ} d } \text{ θ}\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}} \text{ dx }\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
\[\int \sin^5 x\ dx\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int x \sec^2 2x\ dx\]
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
