Advertisements
Advertisements
Question
\[\int\frac{\log \left( \log x \right)}{x} \text{ dx}\]
Sum
Advertisements
Solution
\[\text{ Let I} = \int\frac{\log \left( \log x \right) dx}{x}\]
\[\text{ Putting log x = t}\]
\[ \Rightarrow \frac{1}{x} dx = dt\]
\[ \therefore I = \int 1_{II} \cdot \log _I t \cdot \text{ dt}\]
\[ = \log t\int1\text{ dt }- \int\left\{ \frac{d}{dt}\left( \log t \right)\int1 dt \right\}dt\]
\[ = \log t \cdot t - \int\frac{1}{t} \times t\text{ dt}\]
\[ = \log t \cdot t - \int dt\]
\[ = \log t \cdot t - t + C\]
\[ = t \left( \log t - 1 \right) + C\]
\[ = \log x \left( \text{ log} \left( \log x \right) - 1 \right) + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{\sin^2 x}{1 + \cos x} \text{dx} \]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
\[\ \int\ x \left( 1 - x \right)^{23} dx\]
` ∫ tan^3 x sec^2 x dx `
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
\[\int\frac{\cos x}{\sqrt{4 + \sin^2 x}} dx\]
\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
` ∫ x tan ^2 x dx
\[\int\frac{x^2 \tan^{- 1} x}{1 + x^2} \text{ dx }\]
\[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right) \sqrt{x + 2}}\text{ dx}\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
\[\int \cos^5 x\ dx\]
\[\int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}}\text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int\log \left( x + \sqrt{x^2 + a^2} \right) \text{ dx}\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
