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Question Number 46087 by Saorey last updated on 21/Oct/18

$$\mathrm{please}\mathrm{help}\mathrm{me}!!$$$$\mathrm{calculate}:$$$$\mathrm{I}=\underset{2}{\stackrel{1+{\mathrm{e}}^2}{\int }}\frac{12288\ln (\mathrm{x}-1)}{[{\ln }^{12}(\mathrm{x}-1)+4096](\mathrm{x}-1)}\mathrm{dx}$$$$\mathrm{thanks}!!!$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18

$${\int }_2^{e^2+1}\frac{12288ln(x-1)}{[{ln}^{12}(x-1)+4096](x-1)}dx$$$$t=ln(x-1)dt=\frac{dx}{x-1}$$$${\int }_0^2\frac{12288t}{t^{12}+4096}dt$$$$6144{\int }_0^2\frac{2t}{{\left(t^2\right)}^6+4096}dtk=t^{2}dk=2tdt$$$$6144{\int }_0^4\frac{dk}{k^6+4096}$$$$6144{\int }_0^4\frac{dk}{k^6+4^6}\approx 0$$$$$$$$thecurve\frac{1}{x^6+4^6}concidewithxaxis$$$$soareaunderthecurve$$$${\int }_0^4\frac{dk}{k^6+4^6}\approx 0$$

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18

Commented by Saorey last updated on 21/Oct/18

thank you��

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