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Question Number 144716 by mathdanisur last updated on 28/Jun/21

∫((dx/(e^(2x) +1)))=?

$$\int\left(\frac{{dx}}{{e}^{\mathrm{2}{x}} +\mathrm{1}}\right)=? \\ $$

Answered by imjagoll last updated on 28/Jun/21

 let e^x  = μ ⇒dx = (dμ/e^x ) = (dμ/μ)  I=∫ ((1/(μ^2 +1)))((dμ/μ))  (1/(μ(μ^2 +1))) = (a/μ)+((bμ+c)/(μ^2 +1))  1=(a+b)μ^2 +cμ+a  →μ=0→a=1  μ=1→1=1+b+c+1  →b+c=−1  μ=−1→1=1+b−c+1  →b−c=−1  { ((b=−1)),((c=0)) :}  I=∫((1/μ)−(μ/(μ^2 +1)))dμ  I= ln μ−(1/2)∫((d(μ^2 +1))/(μ^2 +1))  I=ln μ−(1/2)ln (μ^2 +1)+ c  I=ln e^x −(1/2)ln (e^(2x) +1)+c  I=x−(1/2)ln (e^(2x) +1)+c

$$\:\mathrm{let}\:\mathrm{e}^{\mathrm{x}} \:=\:\mu\:\Rightarrow\mathrm{dx}\:=\:\frac{\mathrm{d}\mu}{\mathrm{e}^{\mathrm{x}} }\:=\:\frac{\mathrm{d}\mu}{\mu} \\ $$$$\mathrm{I}=\int\:\left(\frac{\mathrm{1}}{\mu^{\mathrm{2}} +\mathrm{1}}\right)\left(\frac{\mathrm{d}\mu}{\mu}\right) \\ $$$$\frac{\mathrm{1}}{\mu\left(\mu^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{\mathrm{a}}{\mu}+\frac{\mathrm{b}\mu+\mathrm{c}}{\mu^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{1}=\left(\mathrm{a}+\mathrm{b}\right)\mu^{\mathrm{2}} +\mathrm{c}\mu+\mathrm{a} \\ $$$$\rightarrow\mu=\mathrm{0}\rightarrow\mathrm{a}=\mathrm{1} \\ $$$$\mu=\mathrm{1}\rightarrow\mathrm{1}=\mathrm{1}+\mathrm{b}+\mathrm{c}+\mathrm{1} \\ $$$$\rightarrow\mathrm{b}+\mathrm{c}=−\mathrm{1} \\ $$$$\mu=−\mathrm{1}\rightarrow\mathrm{1}=\mathrm{1}+\mathrm{b}−\mathrm{c}+\mathrm{1} \\ $$$$\rightarrow\mathrm{b}−\mathrm{c}=−\mathrm{1}\:\begin{cases}{\mathrm{b}=−\mathrm{1}}\\{\mathrm{c}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{I}=\int\left(\frac{\mathrm{1}}{\mu}−\frac{\mu}{\mu^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{d}\mu \\ $$$$\mathrm{I}=\:\mathrm{ln}\:\mu−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mu^{\mathrm{2}} +\mathrm{1}\right)}{\mu^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{I}=\mathrm{ln}\:\mu−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mu^{\mathrm{2}} +\mathrm{1}\right)+\:\mathrm{c} \\ $$$$\mathrm{I}=\mathrm{ln}\:\mathrm{e}^{\mathrm{x}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{2x}} +\mathrm{1}\right)+\mathrm{c} \\ $$$$\mathrm{I}=\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{2x}} +\mathrm{1}\right)+\mathrm{c}\: \\ $$

Commented by mathdanisur last updated on 28/Jun/21

thanks Sir cool

$${thanks}\:{Sir}\:{cool} \\ $$

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