Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 24670 by A1B1C1D1 last updated on 24/Nov/17

Answered by mrW1 last updated on 24/Nov/17

$$I=\int \sqrt{e^{\lambda x}+k}dx$$$$letu=\sqrt{e^{\lambda x}+k}$$$$du=\frac{e^{\lambda x}\lambda }{2\sqrt{e^{\lambda x}+k}}dx$$$$du=\frac{\lambda }{2}\times \frac{e^{\lambda x}\sqrt{e^{\lambda x}+k}}{(e^{\lambda x}+k)}dx$$$$\Rightarrow \sqrt{e^{\lambda x}+k}dx=\frac{2}{\lambda }\times \frac{e^{\lambda x}+k}{e^{\lambda x}}du=\frac{2}{\lambda }\times \frac{u^2}{u^2-k}du$$$$\Rightarrow I=\frac{2}{\lambda }\int \frac{u^2}{u^2-k}du$$$$=\frac{2}{\lambda }\int \frac{u^2-k+k}{u^2-k}du$$$$=\frac{2}{\lambda }\int (1+\frac{k}{u^2-k})du$$$$=\frac{2}{\lambda }[u+\frac{\sqrt{k}}{2}\ln \frac{u-\sqrt{k}}{u+\sqrt{k}}]+C$$$$=\frac{2}{\lambda }[\sqrt{e^{\lambda x}+k}+\frac{\sqrt{k}}{2}\ln \frac{\sqrt{e^{\lambda x}+k}-\sqrt{k}}{\sqrt{e^{\lambda x}+k}+\sqrt{k}}]+C$$

Commented by A1B1C1D1 last updated on 24/Nov/17

$$\mathrm{Thank}\mathrm{you}.$$

Commented by A1B1C1D1 last updated on 24/Nov/17

$$\mathrm{Thank}\mathrm{you}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com