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Question Number 142717 by mohammad17 last updated on 04/Jun/21

Commented by mohammad17 last updated on 04/Jun/21

$$itisveryhardhowcansolvethis$$

Answered by Ar Brandon last updated on 04/Jun/21

$$\mathrm{I}=\int \frac{2\mathrm{x}+3}{(3\mathrm{x}+4)\sqrt{{\mathrm{x}}^2+2\mathrm{x}+4}}\mathrm{dx}=\frac{2}{3}\int \frac{3\mathrm{x}+\frac{9}{2}}{(3\mathrm{x}+4)\sqrt{{\mathrm{x}}^2+2\mathrm{x}+4}}\mathrm{dx}$$$$=\frac{2}{3}\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+2\mathrm{x}+4}}+\frac{1}{3}\int \frac{\mathrm{dx}}{(3\mathrm{x}+4)\sqrt{{\mathrm{x}}^2+2\mathrm{x}+4}}$$$$\mathrm{J}=\frac{2}{3}\int \frac{\mathrm{dx}}{\sqrt{{(\mathrm{x}+1)}^2+3}}=\frac{2}{3}\mathrm{argsh}\left(\frac{\mathrm{x}+1}{\sqrt{3}}\right)=\frac{2}{3}\ln \mid (\mathrm{x}+1)+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+4}\mid$$$$\mathrm{K}=\frac{1}{3}\int \frac{\mathrm{dx}}{(3\mathrm{x}+4)\sqrt{{\mathrm{x}}^2+2\mathrm{x}+4}},\mathrm{u}=\frac{1}{3\mathrm{x}+4},\mathrm{x}=\frac{1}{3\mathrm{u}}-\frac{4}{3}$$$$=-\frac{1}{3}\cdot \frac{1}{3}\int \frac{\mathrm{u}}{\sqrt{({(\frac{1}{3\mathrm{u}}-\frac{4}{3})}^2+2(\frac{1}{3\mathrm{u}}-\frac{4}{3})+4}}\cdot \frac{\mathrm{du}}{{\mathrm{u}}^2}$$$$=-\frac{1}{9}\int \frac{\mathrm{du}}{\mathrm{u}\sqrt{\frac{1}{{9\mathrm{u}}^2}-\frac{8}{9\mathrm{u}}+\frac{16}{9}+\frac{2}{3\mathrm{u}}-\frac{8}{3}+4}}$$$$=\mp \frac{1}{9}\int \frac{\mathrm{du}}{\sqrt{\frac{1}{9}-\frac{2\mathrm{u}}{9}+\frac{{28\mathrm{u}}^2}{9}}}=\mp \frac{1}{9}\cdot \frac{3}{\sqrt{28}}\int \frac{\mathrm{du}}{\sqrt{{\mathrm{u}}^2-\frac{\mathrm{u}}{14}+\frac{1}{28}}}$$$$=\mp \frac{1}{6\sqrt{7}}\int \frac{\mathrm{du}}{\sqrt{{(\mathrm{u}-\frac{1}{28})}^2+\frac{27}{784}}}=\mp \frac{1}{6\sqrt{7}}\cdot \mathrm{argsh}\left(\frac{28\mathrm{u}-1}{\sqrt{27}}\right)$$$$=\mp \frac{1}{6\sqrt{7}}\ln \mid (\mathrm{u}-\frac{1}{28})+\sqrt{{\mathrm{u}}^2-\frac{\mathrm{u}}{14}+\frac{1}{28}}\mid$$$$=\mp \frac{1}{6\sqrt{7}}\ln \mid (\frac{1}{3\mathrm{x}+4}-\frac{1}{28})+\sqrt{\frac{1}{{(3\mathrm{x}+4)}^2}-\frac{1}{14(3\mathrm{x}+4)}+\frac{1}{28}}\mid +\mathrm{C}$$$$\mathrm{I}=\mathrm{J}+\mathrm{K}$$

Answered by MJS_new last updated on 04/Jun/21

$$\frac{2x+3}{3x+4}=\frac{2}{3}+\frac{1}{3(3x+4)}$$$$\int \frac{2x+3}{(3x+4)\sqrt{x^2+2x+4}}dx=$$$$=\frac{2}{3}\int \frac{dx}{\sqrt{x^2+2x+4}}+\frac{1}{3}\int \frac{dx}{(3x+4)\sqrt{x^2+2x+4}}=$$$$[t=x+1\to dx=dt]$$$$=\frac{2}{3}\int \frac{dt}{\sqrt{t^2+3}}+\frac{1}{3}\int \frac{dt}{(3t+1)\sqrt{t^2+3}}=$$$$[u=\frac{t+\sqrt{t^2+3}}{\sqrt{3}}\to dt=\frac{\sqrt{3(t^2+3)}}{t+\sqrt{t^2+3}}du=\frac{\sqrt{t^2+3}}{u}du]$$$$(\mathrm{usually}\mathrm{you}\mathrm{would}\mathrm{now}\mathrm{substitute}$$$$t=\sqrt{3}\sinh u\mathrm{but}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{like}\mathrm{dealing}$$$$\mathrm{with}\mathrm{hyperbolic}\mathrm{functions})$$$$=\frac{2}{3}\int \frac{du}{u}+\frac{2\sqrt{3}}{27}\int \frac{du}{u^2+\frac{2\sqrt{3}}{9}u-1}=$$$$[v=u+\frac{\sqrt{3}}{9}\to du=dv]$$$$=\frac{2}{3}\ln u+\frac{2\sqrt{3}}{27}\int \frac{dv}{v^2-\frac{28}{27}}=$$$$[w=\frac{3\sqrt{21}}{14}v\to dv=\frac{2\sqrt{21}}{9}dw]$$$$=\frac{2}{3}\ln u+\frac{\sqrt{7}}{21}\int \frac{dw}{w^2-1}=$$$$=\frac{2}{3}\ln u+\frac{\sqrt{7}}{42}\ln \frac{w-1}{w+1}=$$$$...$$$$=\frac{2}{3}\ln (x+1+\sqrt{x^2+2x+4})+\frac{\sqrt{7}}{42}\ln \mid \frac{x-8+2\sqrt{7(x^2+2x+4)}}{3x+4}\mid +C$$

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