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Question Number 96161 by bemath last updated on 30/May/20

$$\underset{1}{\stackrel{4}{\int }}\frac{\mathrm{sech}{}^2\left(\sqrt{x}\right)+\tanh \left(\sqrt{x}\right)}{\sqrt{x}}dx?$$

Commented by bemath last updated on 30/May/20

$$\mathrm{thanks}$$

Answered by bobhans last updated on 30/May/20

$$\mathrm{let}{\mathrm{I}}_1=\underset{1}{\stackrel{4}{\int }}\frac{\mathrm{sech}{}^2\left(\sqrt{x}\right)dx}{\sqrt{x}}=2\underset{1}{\stackrel{4}{\int }}d\left(\tanh \sqrt{x}\right)$$$$=2\tanh \sqrt{x}]{}_2^4=2(\tanh \left(2\right)-\tanh \left(1\right))$$$$=2\{\frac{e^2-e^{-2}}{e^2+e^{-2}}-\frac{e-e^{-1}}{e+e^{-1}}\}=2\{\frac{e^4-1}{e^4+1}-\frac{e^2-1}{e^2+1}\}$$$${\mathrm{let}{\mathrm{I}}_2=\underset{2}{\stackrel{4}{\int }}\frac{\tanh \sqrt{x}}{\sqrt{x}}dx=2\ln \left(\cosh \sqrt{x}\right)]}_1^4$$$$=2\ln \left(\frac{e^2+e^{-2}}{2}\right)-2\ln \left(\frac{\mathrm{e}+{\mathrm{e}}^{-1}}{2}\right)$$$$=2\ln \left(\frac{{\mathrm{e}}^2+{\mathrm{e}}^{-2}}{\mathrm{e}+{\mathrm{e}}^{-1}}\right)=2\ln \left(\frac{{\mathrm{e}}^4+1}{{\mathrm{e}}^3+\mathrm{e}}\right)$$$$\mathrm{then}\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2=2\{\frac{{\mathrm{e}}^4-1}{{\mathrm{e}}^4+1}-\frac{{\mathrm{e}}^2-1}{{\mathrm{e}}^2+1}\}+2\ln \left(\frac{{\mathrm{e}}^4+1}{{\mathrm{e}}^3+\mathrm{e}}\right)$$

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