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Question Number 130226 by stelor last updated on 23/Jan/21

$$\mathrm{please}\mathrm{I}\mathrm{need}\mathrm{help}...$$$$\int \left(\frac{{\mathrm{e}}^x+1}{{\mathrm{e}}^{2\mathrm{x}}+1}\right)dx$$

Answered by Lordose last updated on 23/Jan/21

$$\mathrm{\Omega }\stackrel{\mathrm{u}={\mathrm{e}}^{\mathrm{x}}}{=}\int \frac{\mathrm{u}+1}{\mathrm{u}({\mathrm{u}}^2+1)}\mathrm{du}=\int \frac{1}{{\mathrm{u}}^2+1}\mathrm{du}+\int \frac{1}{\mathrm{u}({\mathrm{u}}^2+1)}\mathrm{du}$$$$\mathrm{\Omega }={\tan }^{-1}\left(\mathrm{u}\right)+\int (\frac{\mathrm{a}}{\mathrm{u}}+\frac{\mathrm{bu}+\mathrm{c}}{{\mathrm{u}}^2+1})\mathrm{du}$$$$1=\mathrm{a}({\mathrm{u}}^2+1)+\mathrm{u}(\mathrm{bu}+\mathrm{c})$$$$\mathrm{a}=1$$$$1=(\mathrm{a}+\mathrm{b}){\mathrm{u}}^2+\mathrm{a}+\mathrm{cu}$$$$\mathrm{b}=-1,\mathrm{c}=0$$$$\mathrm{\Omega }={\tan }^{-1}\left(\mathrm{u}\right)+\int \frac{1}{\mathrm{u}}\mathrm{du}-\int \frac{\mathrm{u}}{{\mathrm{u}}^2+1}\mathrm{du}$$$$\mathrm{\Omega }={\tan }^{-1}\left(\mathrm{u}\right)+\ln \left(\mathrm{u}\right)-\frac{1}{2}\ln ({\mathrm{u}}^2+1)+\mathrm{C}$$$$\mathrm{\Omega }={\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{x}-\frac{1}{2}\ln ({\mathrm{e}}^{2\mathrm{x}}+1)+\mathrm{C}$$

Commented by stelor last updated on 23/Jan/21

$$\mathrm{good}......$$

Answered by EDWIN88 last updated on 23/Jan/21

$$\mathrm{let}{\mathrm{e}}^{\mathrm{x}}=\tan \mathrm{r}\wedge \mathrm{dx}=\frac{\sec {}^2\mathrm{r}}{\tan \mathrm{r}}\mathrm{dr}\Rightarrow \mathrm{I}=\int \frac{(\tan \mathrm{r}+1)}{\sec {}^2\mathrm{r}}\left(\frac{\sec {}^2\mathrm{r}}{\tan \mathrm{r}}\right)\mathrm{dr}$$$$\mathrm{I}=\int \frac{\tan \mathrm{r}+1}{\tan \mathrm{r}}\mathrm{dr}=\mathrm{r}+\int \frac{\cos \mathrm{r}}{\sin \mathrm{r}}\mathrm{dr}$$$$\mathrm{I}=\mathrm{r}+\int \frac{\mathrm{d}\left(\sin \mathrm{r}\right)}{\sin \mathrm{r}}=\mathrm{r}+\ln \left(\sin \mathrm{r}\right)+\mathrm{c}$$$$\mathrm{I}={\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)+\ln \left(\frac{{\mathrm{e}}^{\mathrm{x}}}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}+1}}\right)+\mathrm{c}$$$$\mathrm{I}={\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)+\mathrm{x}-\frac{1}{2}\ln ({\mathrm{e}}^{2\mathrm{x}}+1)+\mathrm{c}$$

Commented by bramlexs22 last updated on 23/Jan/21

$$\mathrm{waw}...\mathrm{elegant}$$

Commented by stelor last updated on 23/Jan/21

$$\mathrm{cool}...$$

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