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Question Number 138024 by Algoritm last updated on 09/Apr/21

Answered by TheSupreme last updated on 09/Apr/21

$$e^x=u$$$$dx=\frac{1}{u}du$$$$\int \sqrt{u^2+4u-1}\frac{du}{u}$$$$(u+2)=z$$$$\int \frac{\sqrt{z^2-5}}{z-2}dz$$$$z^2-5=w^2$$$$dz=\frac{w}{\sqrt{w^2+5}}dw$$$$\int \frac{w^2}{\sqrt{w^2+5}}dw=w\left[\sqrt{w^2+5}\right]-\int \sqrt{w^2+5}dw$$$$w\sqrt{w^2+5}-\frac{1}{2}(\sqrt{w^2+5}w+5{sinh}^{-1}\left(\frac{w}{\sqrt{5}}\right))+c$$$$\frac{\sqrt{z^2-5}}{2}z+5{sinh}^{-1}\left(\frac{\sqrt{z^2-5}}{\sqrt{5}}\right)+c$$$$\frac{\sqrt{u^2+4u-1}}{2}(u+2)+5{sinh}^{-1}\left(\frac{\sqrt{u^2+4u-1}}{\sqrt{5}}\right)+c$$$$\frac{\sqrt{e^{2x}+4e^x-1}}{2}(e^x+2)+5{sinh}^{-1}\left(\frac{\sqrt{e^{2x}+4e^x-1}}{\sqrt{5}}\right)+c$$

Answered by MJS_new last updated on 09/Apr/21

$$\int \sqrt{{\mathrm{e}}^{2x}+{4\mathrm{e}}^x-1}dx=\sqrt{{({\mathrm{e}}^x+2)}^2-5}dx=$$$$[t=\frac{{\mathrm{e}}^x+2+\sqrt{{\mathrm{e}}^{2x}+{4\mathrm{e}}^x-1}}{\sqrt{5}}\to dx=\frac{\sqrt{5({\mathrm{e}}^{2x}+{4\mathrm{e}}^x-1)}}{{\mathrm{e}}^x+2+\sqrt{{\mathrm{e}}^{2x}+{4\mathrm{e}}^x-1}}]$$$$=\frac{5}{2}\int \frac{{(t^2-1)}^2}{t^2(\sqrt{5}{t}^2-4t+\sqrt{5})}dt=$$$$=\int (\frac{\sqrt{5}}{2t^2}+\frac{2}{t}+\frac{\sqrt{5}}{2}-\frac{2}{\sqrt{5}{t}^2-4t+\sqrt{5}})dt=$$$$=-\frac{\sqrt{5}}{2t}+2\ln t+\frac{\sqrt{5}t}{2}-2\arctan (\sqrt{5}t-2)=$$$$=\sqrt{{\mathrm{e}}^{2x}+{4\mathrm{e}}^x-1}+2\ln ({\mathrm{e}}^x+2+\sqrt{{\mathrm{e}}^{2x}+{4\mathrm{e}}^x-1})-2\arctan ({\mathrm{e}}^x+\sqrt{{\mathrm{e}}^{2x}+{4\mathrm{e}}^x-1})+C$$

Answered by Mathspace last updated on 09/Apr/21

$$I=\int \sqrt{e^{2x}+4e^x-1}dx$$$$I{=}_{e^x=t}\int \sqrt{t^2+4t-1}\frac{dt}{t}$$$$=\int \frac{\sqrt{t^2+4t+4-5}}{t}dt$$$$=\int \frac{\sqrt{{(t+2)}^2-5}}{t}dt$$$${=}_{t+2=\sqrt{5}ch\left(u\right)}\int \frac{\sqrt{5}shu}{(\sqrt{5}chu-2)}\sqrt{5}chudu$$$$=5\int \frac{{sh}^{2}u}{\sqrt{5}chu-2}du$$$$=5\int \frac{\frac{ch\left(2u\right)-1}{2}}{\sqrt{5}chu-2}du$$$$=\frac{5}{2}\int \frac{ch\left(2u\right)}{\sqrt{5}chu-2}du$$$$=\frac{5}{4}\int \frac{e^{2u}-e^{-2u}}{\sqrt{5}\frac{e^u+e^{-u}}{2}-2}du$$$$=\frac{5}{2}\int \frac{e^{2u}-e^{-2u}}{\sqrt{5}{e}^u+\sqrt{5}{e}^{-u}-4}du$$$${=}_{e^u=y}\frac{5}{2}\int \frac{y^2-y^{-2}}{\sqrt{5}y+\sqrt{5}{y}^{-1}-4}\frac{dy}{y}$$$$=\frac{5}{2}\int \frac{y^2-y^{-2}}{\sqrt{5}{y}^2+\sqrt{5}-4y}dy$$$$=\frac{5}{2}\int \frac{y^4-1}{y^2(\sqrt{5}{y}^2-4y+\sqrt{5})}dy$$$$letdecomposeF\left(y\right)=\frac{y^4-1}{y^2(\sqrt{5}{y}^2-4y+\sqrt{5})}$$$$F\left(y\right)=\frac{1}{\sqrt{5}}+\frac{a}{y}+\frac{b}{y^2}+\frac{cy+d}{\sqrt{5}{y}^2-4y+\sqrt{5}}$$$$...becpntinued...$$$$=$$

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