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Question Number 73030 by Tanmay chaudhury last updated on 05/Nov/19

Answered by Tanmay chaudhury last updated on 05/Nov/19

$${\int }_0^4\frac{ln2}{lnx}-\frac{ln2\times ln2}{lnx\times lnx\times ln2}dx$$$$ln2{\int }_2^4\frac{1}{lnx}-\frac{1}{{\left(lnx\right)}^2}dx$$$$ln2{\int }_2^4\frac{lnx-1}{{\left(lnx\right)}^2}dx$$$$ln2{\int }_2^4\frac{lnx.\frac{dx}{dx}-x.\frac{d}{dx}\left(lnx\right)}{{\left(lnx\right)}^2}d\stackrel{}{x}$$$$ln2{\int }_2^4\frac{d}{dx}\left(\frac{x}{lnx}\right)dx$$$$ln2\times \mid \frac{x}{lnx}{\mid }_2^4$$$$ln2\times (\frac{4}{ln4}-\frac{2}{ln2})$$$$ln2\times (\frac{4}{2ln2}-\frac{2}{ln2})=0answer$$

Answered by mind is power last updated on 05/Nov/19

$${\int }_2^4({\log }_{\mathrm{x}}\left(2\right)-\frac{{\left({\log }_{\mathrm{x}}\left(2\right)\right)}^2}{\ln \left(2\right)})\mathrm{dx}$$$$=\int \frac{\ln \left(2\right)}{\ln \left(\mathrm{x}\right)}-\frac{\ln \left(2\right)}{{\ln }^2\left(\mathrm{x}\right)}\mathrm{dx}$$$$=\ln \left(2\right)\int \frac{\ln \left(\mathrm{x}\right)-1}{{\ln }^2\left(\mathrm{x}\right)}\mathrm{dx}$$$$=\ln \left(2\right)\int \mathrm{d}\left(\frac{\mathrm{x}}{\ln \left(\mathrm{x}\right)}\right)=\ln 2{\left[\frac{\mathrm{x}}{\ln \left(\mathrm{x}\right)}\right]}_2^4=\frac{4\ln \left(2\right)}{\ln \left(4\right)}-\frac{2\ln \left(2\right)}{\ln \left(2\right)}=2-2=0$$

Commented by Tanmay chaudhury last updated on 05/Nov/19

$$thankyousir...$$

Commented by mind is power last updated on 05/Nov/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$$$$$

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