0-1-dx-e-x-e-x-tan-1-e-pi-4-

Question Number 53386 by gunawan last updated on 21/Jan/19

$$\underset{0}{\stackrel{1}{\int }}\frac{dx}{e^x+e^{-x}}={\tan }^{-1}e-\frac{\pi }{4}$$

Commented by Tinkutara last updated on 21/Jan/19

$$=\int \frac{e^{x}dx}{e^{2x}+1}$$$$Put{e}^x=tandintegrate$$

Commented by ajfour last updated on 21/Jan/19

$$howwasyourMainsexam?$$

Commented by gunawan last updated on 21/Jan/19

$$\mathrm{Nice}\mathrm{and}\mathrm{clearly}\mathrm{Sir}$$

Commented by maxmathsup by imad last updated on 21/Jan/19

$$letI={\int }_0^1\frac{dx}{e^x+e^{-x}}\Rightarrow I{=}_{e^x=t}{\int }_1^e\frac{1}{t+t^{-1}}\frac{dt}{t}={\int }_1^e\frac{dt}{t^2+1}$$$$={\left[arctan\left(t\right)\right]}_1^e=arctan\left(e\right)-arctan\left(1\right)=arctan\left(e\right)-\frac{\pi }{4}.$$

Commented by Tinkutara last updated on 24/Jan/19

@ajfour Sir My Mains went well above my nervous expectations… I topped in my district Thanks for your continuous support sir And thanks to this wonderful platform… ��

Commented by ajfour last updated on 24/Jan/19

$$Itsallamatterofyourcuriosity$$$$andwillingnesstolearn.Itoo,$$$$hadenjoyedsolvingyourdoubts!$$

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