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Question Number 185181 by Noorzai last updated on 18/Jan/23

Commented by Shrinava last updated on 18/Jan/23

$$\ln \left(\ln \left(\ln \left(\ln \left(\ln \left(\mathrm{lnx}\right)\right)\right)\right)\right)+\mathbb{C}$$

Commented by Noorzai last updated on 18/Jan/23

$$?$$

Commented by Gazella thomsonii last updated on 18/Jan/23

$$\mathrm{what}\mathrm{the}....$$

Answered by SEKRET last updated on 19/Jan/23

$$\left[\begin{array}{c}\mathbf{ln}\left(\mathbf{x}\right)=\mathbf{a}\\ \frac{1}{\mathbf{x}}\mathbf{dx}=\mathbf{da}\end{array}\right]$$$$\int \frac{\frac{\frac{\frac{\frac{\frac{\frac{1}{\mathbf{a}}}{\mathbf{ln}\left(\mathbf{a}\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{a}\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{a}\right)\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{a}\right)\right)\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{a}\right)\right)\right)\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{a}\right)\right)\right)\right)\right)\right)}\mathbf{da}\left[\begin{array}{c}\mathbf{ln}\left(\mathbf{a}\right)=\mathbf{b}\\ \frac{1}{\mathbf{a}}\mathbf{da}=\mathbf{db}\end{array}\right]$$$$\int \frac{\frac{\frac{\frac{\frac{\frac{1}{\mathbf{b}}}{\mathbf{ln}\left(\mathbf{b}\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{b}\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{b}\right)\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{b}\right)\right)\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{b}\right)\right)\right)\right)\right)}\mathbf{db}\left[\begin{array}{c}\mathbf{ln}\left(\mathbf{b}\right)=\mathbf{c}\\ \frac{1}{\mathbf{b}}\mathbf{db}=\mathbf{dc}\end{array}\right]$$$$\int \frac{\frac{\frac{\frac{\frac{1}{\mathbf{c}}}{\mathbf{ln}\left(\mathbf{c}\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{c}\right)\right))}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{c}\right)\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{c}\right)\right)\right)\right)}\mathbf{dc}\left[\begin{array}{c}\mathbf{ln}\left(\mathbf{c}\right)=\mathbf{e}\\ \frac{1}{\mathbf{c}}\mathbf{dc}=\mathbf{de}\end{array}\right]$$$$\int \frac{\frac{\frac{\frac{1}{\mathbf{e}}}{\mathbf{ln}\left(\mathbf{e}\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{e}\right)\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{e}\right)\right)\right)}\mathbf{de}\left[\begin{array}{c}\mathbf{ln}\left(\mathbf{e}\right)=\mathbf{f}\\ \frac{1}{\mathbf{e}}\mathbf{de}=\mathbf{df}\end{array}\right]$$$$\int \frac{\frac{\frac{1}{\mathbf{f}}}{\mathbf{ln}\left(\mathbf{f}\right)}}{\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{f}\right)\right)}\mathbf{df}\left[\begin{array}{c}\mathbf{ln}\left(\mathbf{f}\right)=\mathbf{g}\\ \frac{1}{\mathbf{f}}\mathbf{df}=\mathbf{dg}\end{array}\right]$$$$\int \frac{\frac{1}{\mathbf{g}}}{\mathbf{ln}\left(\mathbf{g}\right)}\mathbf{dg}\left[\begin{array}{c}\mathbf{ln}\left(\mathbf{g}\right)=\mathbf{h}\\ \frac{1}{\mathbf{g}}\mathbf{dg}=\mathbf{h}\end{array}\right]$$$$\int \frac{1}{\mathbf{h}}\mathbf{dh}=\mathbf{ln}\left(\mathbf{h}\right)+\mathbf{c}$$$$\mathbf{h}=\mathbf{ln}\left(\mathbf{g}\right)=\mathbf{ln}\left(\mathbf{lnf}\right))=\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{e}\right)\right)\right)=$$$$=\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{c}\right)\right)\right)\right)=\mathbf{ln}(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{x}\right)\right)\right)\right)\right)\right)$$$$\mathbf{answer}$$$$\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{x}\right)\right)\right)\right)\right)\right)\right)+\mathbf{C}$$$$\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{L}\mathit{A}\mathit{Z}\mathit{I}\mathit{Z}\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{V}\mathit{A}\mathit{L}\mathit{I}\mathit{Y}\mathit{E}\mathit{V}$$$$$$$$$$

Answered by aba last updated on 18/Jan/23

$$\ln \mid \ln (\ln (\ln \left(\ln \left(\ln \left(\ln \left(\ln \left(\mathrm{x}\right)\right)\right)\right)\right)\mid +\mathrm{k}/\mathrm{k}\in \mathrm{R}$$

Commented by Noorzai last updated on 18/Jan/23

$$Beexplainedyou$$

Commented by aba last updated on 18/Jan/23

$$\int \frac{{\mathrm{u}}^{{}^{\prime }}}{\mathrm{u}}\mathrm{du}=\ln \mid \mathrm{u}\mid +\mathbb{k}$$

Commented by Noorzai last updated on 19/Jan/23

$$thanks$$

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