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x-cosh-x-sinh-x-2-dx-




Question Number 132473 by physicstutes last updated on 14/Feb/21
∫ ((x cosh x)/((sinh x)^2 )) dx
$$\int\:\frac{{x}\:\mathrm{cosh}\:{x}}{\left(\mathrm{sinh}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$
Answered by mathmax by abdo last updated on 14/Feb/21
I=∫ ((xchx)/(sh^2 x))dx  by parts  u^′  =((chx)/(sh^2 x)) and v=x ⇒  I =−(x/(shx))−∫  (−(1/(shx)))dx =−(x/(shx)) +∫  (dx/(shx)) we have   ∫ (dx/(shx)) =2∫ (dx/(e^x −e^(−x) )) =_(e^x  =t)   2∫  (dt/(t(t−t^(−1) ))) =∫ ((2dt)/(t^2 −1))  =∫((1/(t−1))−(1/(t+1)))dt =ln∣((t−1)/(t+1))∣ +C =ln∣((e^x −1)/(e^x +1))∣+C ⇒  I =−(x/(shx)) +ln∣((e^x −1)/(e^x  +1))∣ +C
$$\mathrm{I}=\int\:\frac{\mathrm{xchx}}{\mathrm{sh}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\:\mathrm{u}^{'} \:=\frac{\mathrm{chx}}{\mathrm{sh}^{\mathrm{2}} \mathrm{x}}\:\mathrm{and}\:\mathrm{v}=\mathrm{x}\:\Rightarrow \\ $$$$\mathrm{I}\:=−\frac{\mathrm{x}}{\mathrm{shx}}−\int\:\:\left(−\frac{\mathrm{1}}{\mathrm{shx}}\right)\mathrm{dx}\:=−\frac{\mathrm{x}}{\mathrm{shx}}\:+\int\:\:\frac{\mathrm{dx}}{\mathrm{shx}}\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\int\:\frac{\mathrm{dx}}{\mathrm{shx}}\:=\mathrm{2}\int\:\frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }\:=_{\mathrm{e}^{\mathrm{x}} \:=\mathrm{t}} \:\:\mathrm{2}\int\:\:\frac{\mathrm{dt}}{\mathrm{t}\left(\mathrm{t}−\mathrm{t}^{−\mathrm{1}} \right)}\:=\int\:\frac{\mathrm{2dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)\mathrm{dt}\:=\mathrm{ln}\mid\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\mid\:+\mathrm{C}\:=\mathrm{ln}\mid\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}\mid+\mathrm{C}\:\Rightarrow \\ $$$$\mathrm{I}\:=−\frac{\mathrm{x}}{\mathrm{shx}}\:+\mathrm{ln}\mid\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{x}} \:+\mathrm{1}}\mid\:+\mathrm{C} \\ $$

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