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e-ax-b-dx-




Question Number 102043 by Dwaipayan Shikari last updated on 06/Jul/20
∫e^(√(ax+b)) dx
$$\int{e}^{\sqrt{{ax}+{b}}} {dx} \\ $$
Commented by Dwaipayan Shikari last updated on 06/Jul/20
(2/a)∫e^u udu             {suppose ax+b=u^2   (2/a)ue^u −(2/a)e^u =(2/a)e^(√(ax+b)) ((√(ax+b))−1)
$$\frac{\mathrm{2}}{{a}}\int{e}^{{u}} {udu}\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{suppose}\:{ax}+{b}={u}^{\mathrm{2}} \right. \\ $$$$\frac{\mathrm{2}}{{a}}{ue}^{{u}} −\frac{\mathrm{2}}{{a}}{e}^{{u}} =\frac{\mathrm{2}}{{a}}{e}^{\sqrt{{ax}+{b}}} \left(\sqrt{{ax}+{b}}−\mathrm{1}\right) \\ $$$$ \\ $$

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