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Question Number 76469 by kaivan.ahmadi last updated on 27/Dec/19

$$\int cothxdx$$

Commented by mathmax by abdo last updated on 27/Dec/19

$$\int coth\left(x\right)dx=\int \frac{ch\left(x\right)}{sh\left(x\right)}dx=\int \frac{e^x+e^{-x}}{e^x-e^{-x}}dx$$$${=}_{e^x=t}\int \frac{t+t^{-1}}{t-t^{-1}}\frac{dt}{t}=\int \frac{t+t^{-1}}{t^2-1}dt=\int \frac{t^2+1}{t(t^2-1)}dt$$$$letdecomposeF\left(t\right)=\frac{t^2+1}{t(t^2-1)}=\frac{t^2+1}{t(t-1)(t+1)}\Rightarrow F\left(t\right)=\frac{a}{t}+\frac{b}{t-1}+\frac{c}{t+1}$$$$a=tF\left(t\right){\mid }_{t=0}=-1$$$$b=(t-1)F\left(t\right){\mid }_{t=1}=1$$$$c=(t+1)F\left(t\right){\mid }_{t=-1}=\frac{2}{2}=1\Rightarrow F\left(t\right)=-\frac{1}{t}+\frac{1}{t-1}+\frac{1}{t+1}\Rightarrow$$$$\int F\left(t\right)dt=ln\mid t+1\mid +ln\mid t-1\mid -ln\mid t\mid +c$$$$=ln(e^x+1)+ln(e^x-1)-x+c$$$$=ln(e^{2x}-1)-x+c\Rightarrow \int coth\left(x\right)dx=ln(e^{2x}-1)-x+c$$

Commented by mathmax by abdo last updated on 27/Dec/19

$$forgive\int coth\left(x\right)dx=ln\mid e^{2x}-1\mid -x+c$$

Commented by kaivan.ahmadi last updated on 29/Dec/19

$$thanksir$$

Answered by Rio Michael last updated on 27/Dec/19

$$\int cothxdx=\int \frac{coshx}{sinhx}dx$$$$letu=sinhx\Rightarrow du=coshxdx$$$$\Rightarrow \int \frac{coshx}{sinhx}dx=\int \frac{coshx}{u}\frac{du}{coshx}$$$$=\int \frac{1}{u}du=ln\mid u\mid +A$$$$=ln\mid sinhx\mid +A$$

Commented by kaivan.ahmadi last updated on 29/Dec/19

$$thanku$$

Answered by Henri Boucatchou last updated on 28/Dec/19

$$=ln(e^x-e^{-x})+C^{st}$$

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