Advertisements
Advertisements
Question
\[\int\frac{1}{ x \text{log x } \text{log }\left( \text{log x }\right)} dx\]
Sum
Advertisements
Solution
` 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
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{\left( 1 + x \right)^3}{\sqrt{x}} dx\]
\[\int\left\{ x^2 + e^{\log x}+ \left( \frac{e}{2} \right)^x \right\} dx\]
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}} dx\]
\[\int \left( \tan x + \cot x \right)^2 dx\]
\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]
\[\int\frac{1}{1 - \sin x} dx\]
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
\[\int \cos^2 \text{nx dx}\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} dx\]
\[\int\frac{x^2}{\sqrt{x - 1}} dx\]
\[\int \sin^4 x \cos^3 x \text{ dx }\]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{1}{4 x^2 + 12x + 5} dx\]
\[\int\frac{1}{x^2 + 6x + 13} dx\]
\[\int\frac{e^x}{\left( 1 + e^x \right)\left( 2 + e^x \right)} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
\[\int\frac{2x - 3}{x^2 + 6x + 13} dx\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int x^3 \cos x^2 dx\]
\[\int x \sin x \cos x\ dx\]
\[\int e^\sqrt{x} \text{ dx }\]
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} dx\]
\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \sqrt{x + 2}}\text{ dx}\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int \tan^5 x\ dx\]
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int x \sec^2 2x\ dx\]
