Question Number 24670 by A1B1C1D1 last updated on 24/Nov/17
Answered by mrW1 last updated on 24/Nov/17
![I=∫(√(e^(λx) +k)) dx let u=(√(e^(λx) +k)) du=((e^(λx) λ)/(2(√(e^(λx) +k)))) dx du=(λ/2)×((e^(λx) (√(e^(λx) +k)))/((e^(λx) +k))) dx ⇒(√(e^(λx) +k)) dx=(2/λ)×((e^(λx) +k)/e^(λx) ) du=(2/λ)×(u^2 /(u^2 −k)) du ⇒I=(2/λ)∫(u^2 /(u^2 −k)) du =(2/λ)∫((u^2 −k+k)/(u^2 −k)) du =(2/λ)∫(1+(k/(u^2 −k))) du =(2/λ)[u+((√k)/2)ln ((u−(√k))/(u+(√k)))]+C =(2/λ)[(√(e^(λx) +k))+((√k)/2)ln (((√(e^(λx) +k))−(√k))/( (√(e^(λx) +k))+(√k)))]+C](https://www.tinkutara.com/question/Q24674.png)
$${I}=\int\sqrt{{e}^{\lambda{x}} +{k}}\:{dx} \\ $$$${let}\:{u}=\sqrt{{e}^{\lambda{x}} +{k}} \\ $$$${du}=\frac{{e}^{\lambda{x}} \lambda}{\mathrm{2}\sqrt{{e}^{\lambda{x}} +{k}}}\:{dx} \\ $$$${du}=\frac{\lambda}{\mathrm{2}}×\frac{{e}^{\lambda{x}} \sqrt{{e}^{\lambda{x}} +{k}}}{\left({e}^{\lambda{x}} +{k}\right)}\:{dx} \\ $$$$\Rightarrow\sqrt{{e}^{\lambda{x}} +{k}}\:{dx}=\frac{\mathrm{2}}{\lambda}×\frac{{e}^{\lambda{x}} +{k}}{{e}^{\lambda{x}} }\:{du}=\frac{\mathrm{2}}{\lambda}×\frac{{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} −{k}}\:{du} \\ $$$$\Rightarrow{I}=\frac{\mathrm{2}}{\lambda}\int\frac{{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} −{k}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\lambda}\int\frac{{u}^{\mathrm{2}} −{k}+{k}}{{u}^{\mathrm{2}} −{k}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\lambda}\int\left(\mathrm{1}+\frac{{k}}{{u}^{\mathrm{2}} −{k}}\right)\:{du} \\ $$$$=\frac{\mathrm{2}}{\lambda}\left[{u}+\frac{\sqrt{{k}}}{\mathrm{2}}\mathrm{ln}\:\frac{{u}−\sqrt{{k}}}{{u}+\sqrt{{k}}}\right]+{C} \\ $$$$=\frac{\mathrm{2}}{\lambda}\left[\sqrt{{e}^{\lambda{x}} +{k}}+\frac{\sqrt{{k}}}{\mathrm{2}}\mathrm{ln}\:\frac{\sqrt{{e}^{\lambda{x}} +{k}}−\sqrt{{k}}}{\:\sqrt{{e}^{\lambda{x}} +{k}}+\sqrt{{k}}}\right]+{C} \\ $$
Commented by A1B1C1D1 last updated on 24/Nov/17
$$\mathrm{Thank}\:\mathrm{you}. \\ $$
Commented by A1B1C1D1 last updated on 24/Nov/17
$$\mathrm{Thank}\:\mathrm{you}. \\ $$