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Question Number 84884 by bshahid010@gmail.com last updated on 17/Mar/20

Commented by mathmax by abdo last updated on 17/Mar/20

$$A=\int \frac{dx}{(2x-1)\sqrt{x^2+x+3}}changement2x-1=tgivex=\frac{t+1}{2}$$$$A=\int \frac{dt}{2t\sqrt{\frac{{(t+1)}^2}{4}+\frac{t+1}{2}+3}}=\int \frac{dt}{2t\sqrt{\frac{t^2+2t+1+2t+2+12}{4}}}$$$$=\int \frac{dt}{t\sqrt{t^2+4t+15}}=\int \frac{dt}{t\sqrt{{(t+2)}^2+11}}$$$${=}_{t+2=\sqrt{11}sh\left(u\right)}\int \frac{\sqrt{11}ch\left(u\right)dt}{(\sqrt{11}sh\left(u\right)-2)\sqrt{11}ch\left(u\right)}$$$$=\int \frac{du}{\sqrt{11}\times \frac{e^u-e^{-u}}{2}-2}=\int \frac{2du}{\sqrt{11}{e}^u-\sqrt{11}{e}^{-u}-4}$$$${=}_{e^u=\alpha }\int \frac{2d\alpha }{\alpha (\sqrt{11}\alpha -\sqrt{11}{\alpha }^{-1}-4)}=2\int \frac{d\alpha }{\sqrt{11}{\alpha }^2-4\alpha -\sqrt{11}}$$$${\mathrm{\Delta }}^{{}^{\prime }}=4+11=15\Rightarrow {\alpha }_1=\frac{2+\sqrt{15}}{\sqrt{11}}and{\alpha }_2=\frac{2-\sqrt{15}}{\sqrt{11}}\Rightarrow$$$$A=2\int \frac{d\alpha }{\sqrt{11}(\alpha -{\alpha }_1)(\alpha -{\alpha }_2)}$$$$=\frac{2}{\sqrt{11}\times \frac{2\sqrt{15}}{\sqrt{11}}}\int (\frac{1}{\alpha -{\alpha }_1}-\frac{1}{\alpha -{\alpha }_2})d\alpha =\frac{1}{\sqrt{15}}ln\mid \frac{\alpha -{\alpha }_1}{\alpha -{\alpha }_2}\mid +C$$$$=\frac{1}{\sqrt{15}}ln\mid \frac{e^u-{\alpha }_1}{e^u-{\alpha }_2}\mid +Cwehaveu=argsh\left(\frac{t+2}{\sqrt{11}}\right)=argsh\left(\frac{2x-1+2}{\sqrt{11}}\right)$$$$=ln(\frac{2x+1}{\sqrt{11}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{11}}\right)}^2})\Rightarrow$$$$A=\frac{1}{\sqrt{15}}ln\mid \frac{\frac{2x+1}{\sqrt{11}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{11}}\right)}^2}-\frac{2+\sqrt{15}}{\sqrt{11}}}{\frac{2x+1}{\sqrt{11}}+\sqrt{1+{\left(\frac{2x+1}{\sqrt{11}}\right)}^2-\frac{2-\sqrt{15}}{\sqrt{11}}}}\mid +C$$$$$$$$$$

Answered by MJS last updated on 17/Mar/20

$$\int \frac{dx}{(2x-1)\sqrt{x^2+x+3}}=$$$$[t=\sqrt{x^2+x+3}\to \frac{2\sqrt{x^2+x+3}}{2x+1}dt]$$$$=\int \frac{dt}{2t^2-\frac{11}{2}-\sqrt{4t^2-11}}=$$$$[t=\frac{\sqrt{11}}{2}\cosh \ln u=\frac{\sqrt{11}(u^2+1)}{4u}\iff u=\frac{2t+\sqrt{4t^2-11}}{\sqrt{11}}\to dt=\frac{\sqrt{11(4t^2-11)}}{2(2t+\sqrt{4t^2-11})}du]$$$$=\frac{2}{\sqrt{11}}\int \frac{du}{u^2-\frac{4}{\sqrt{11}}u-1}=$$$$=\frac{1}{\sqrt{15}}\int \frac{du}{u-\frac{2+\sqrt{15}}{\sqrt{11}}}-\frac{1}{\sqrt{15}}\int \frac{du}{u-\frac{2-\sqrt{15}}{\sqrt{11}}}$$$$\mathrm{and}\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}$$

Answered by TANMAY PANACEA last updated on 17/Mar/20

$$2x-1=\frac{1}{t}\to 2dx=\frac{-1}{t^2}dt$$$$2x=(1+\frac{1}{t})\to x=\frac{1}{2}(1+\frac{1}{t})$$$$dx=\frac{-dt}{2t^2}$$$$\int \frac{-dt}{2t^2\times \frac{1}{t}\sqrt{\frac{1}{4}{(1+\frac{1}{t})}^2+\frac{1}{2}(1+\frac{1}{t})+3}}$$$$\int \frac{-dt}{2t\sqrt{\frac{1}{4}(1+\frac{2}{t}+\frac{1}{t^2})+\frac{1}{2}+\frac{1}{2t}+3}}$$$$\frac{-1}{2}\int \frac{dt}{t\sqrt{\frac{1}{4}+\frac{1}{2t}+\frac{1}{4t^2}+\frac{1}{2}+\frac{1}{2t}+3}}$$$$\frac{-1}{2}\int \frac{dt}{t\sqrt{\frac{t^2+2t+1+2t^2+2t+12t^2}{4t^2}}}$$$$\frac{-1}{2}\int \frac{2tdt}{t\sqrt{15t^2+4t+1}}$$$$-\int \frac{dt}{\sqrt{15(t^2+\frac{4t}{15}+\frac{1}{15})}}$$$$\frac{-1}{\sqrt{15}}\int \frac{dt}{\sqrt{{(t+\frac{2}{15})}^2+\frac{1}{15}-\frac{4}{{15}^2}}}$$$$\frac{-1}{\sqrt{15}}\int \frac{dt}{\sqrt{{(t+\frac{2}{15})}^2+{\left(\frac{\sqrt{11}}{15}\right)}^2}}$$$$\frac{-1}{\sqrt{15}}\times ln\{(t+\frac{2}{15})+\sqrt{{(t+\frac{2}{15})}^2+{\left(\frac{\sqrt{11}}{15}\right)}^2}\}+C$$$$nowputt=\frac{1}{2x-1}$$

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