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dy-ae-y-1-




Question Number 161381 by mkam last updated on 17/Dec/21
∫ (dy/( (√(ae^y +1))))
$$\int\:\frac{{dy}}{\:\sqrt{{ae}^{{y}} +\mathrm{1}}} \\ $$
Answered by MohammadAzad last updated on 17/Dec/21
u=e^y   du=e^y dy  ∫(dy/( (√(ae^y +1))))=∫(du/(u(√(au+1))))  =ln∣(((√(au+1))−1)/( (√(au+1))+1))∣+C  =ln∣(((√(ae^y +1))−1)/( (√(ae^y +1))+1))∣+C
$${u}={e}^{{y}} \\ $$$${du}={e}^{{y}} {dy} \\ $$$$\int\frac{{dy}}{\:\sqrt{{ae}^{{y}} +\mathrm{1}}}=\int\frac{{du}}{{u}\sqrt{{au}+\mathrm{1}}} \\ $$$$=\mathrm{ln}\mid\frac{\sqrt{{au}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{au}+\mathrm{1}}+\mathrm{1}}\mid+{C} \\ $$$$=\mathrm{ln}\mid\frac{\sqrt{{ae}^{{y}} +\mathrm{1}}−\mathrm{1}}{\:\sqrt{{ae}^{{y}} +\mathrm{1}}+\mathrm{1}}\mid+{C} \\ $$

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