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Question Number 154875 by mathdanisur last updated on 22/Sep/21

Answered by mindispower last updated on 23/Sep/21

$$u=log(x+\frac{5}{x})$$$$\mathrm{\Omega }\iff \int e^{2u}sin\left(u\right)du$$

Commented by mathdanisur last updated on 23/Sep/21

$$\mathrm{thanks}\mathbf{S}\mathrm{er},\mathrm{but}\mathrm{how}$$

Commented by mindispower last updated on 23/Sep/21

$$u=ln(x+\frac{5}{x})\Rightarrow du=\frac{1-\frac{5}{x^2}}{x+\frac{5}{x}}dx\Rightarrow (1-\frac{5}{x^2})dx={ue}^{u}du$$$$\int {(x+\frac{5}{x})}_{=e^u}sin\left(ln(x+\frac{5}{x})\right).(1-\frac{5}{x^2}){dx}_{=e^{u}du}$$$$=\int e^{u}sin\left(u\right).e^{u}du=\int e^{2u}sin\left(u\right)du$$

Commented by mathdanisur last updated on 23/Sep/21

$$\mathrm{very}\mathrm{nise}\mathbf{S}\mathrm{er},\mathrm{thank}\mathrm{you}$$

Commented by mindispower last updated on 23/Sep/21

$$withePleasur$$

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