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Question Number 86302 by john santu last updated on 28/Mar/20

$$\int \frac{dx}{(x^2+1)\sqrt{x^2+4}}=?$$

Commented by abdomathmax last updated on 28/Mar/20

$$I=\int \frac{dx}{(x^2+1)\sqrt{x^2+4}}changementx=2sh\left(t\right)give$$$$I=\int \frac{2ch\left(t\right)dt}{(1+4{sh}^{2}t)2ch\left(t\right)}=\int \frac{dt}{1+4\times \frac{ch\left(2t\right)-1}{2}}$$$$=\int \frac{2dt}{2+4ch\left(2t\right)-4}=\int \frac{dt}{2ch\left(2t\right)-1}$$$$=\int \frac{dt}{2\times \frac{e^{2t}+e^{-2t}}{2}-1}=\int \frac{dt}{e^{2t}+e^{-2t}-1}$$$${=}_{e^t=z}\int \frac{dz}{z(z^2+z^{-2}-1)}=\int \frac{dz}{z^3-z^{-1}-z}$$$$=\int \frac{zdz}{z^4-1-z^2}=\int \frac{zdz}{z^4-z^2-1}letdecompose$$$$F\left(z\right)=\frac{z}{z^4-z^2-1}$$$$z^4-z^2-1=0\Rightarrow t^2-t-1=0(t=z^2)$$$$\mathrm{\Delta }=1+4=5\Rightarrow t_1=1+\sqrt{5}and{t}_2=1-\sqrt{5}$$$$\Rightarrow F\left(z\right)=\frac{z}{(z^2-t_1)(z^2-t_2)}=\frac{z}{2\sqrt{5}}(\frac{1}{z^2-t_1}-\frac{1}{z^2-t_2})$$$$\Rightarrow \int F\left(z\right)dz=\frac{1}{2\sqrt{5}}ln\left(\frac{z^2-t_1}{z^2-t_2}\right)+C$$$$=\frac{1}{2\sqrt{5}}ln\left(\frac{e^{2t}-t_1}{e^{2t}-t_2}\right)+C$$$$t=argsh\left(\frac{x}{2}\right)=ln(\frac{x}{2}+\sqrt{1+\frac{x^2}{4}})\Rightarrow$$$$e^{2t}={(\frac{x}{2}+\sqrt{1+\frac{x^2}{4}})}^2\Rightarrow$$$$I=\frac{1}{2\sqrt{5}}ln\left(\frac{{(\frac{x}{2}+\sqrt{1+\frac{x^2}{2}})}^2-1-\sqrt{5}}{{(\frac{x}{2}+\sqrt{1+\frac{x^2}{4}})}^2-1+\sqrt{5}}\right)+C$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$$x=\frac{1}{t}\to dx=\frac{-dt}{t^2}$$$$\int \frac{-dt}{t^2(\frac{1}{t^2}+1)\sqrt{\frac{1}{t^2}+4}}$$$$\int \frac{-tdt}{(1+t^2)\sqrt{1+4t^2}}$$$$t^2=k\to dk=2tdt$$$$\int \frac{-dk}{2(1+k)\sqrt{1+4k}}$$$$\frac{-1}{2}\int \frac{dk}{(1+k)\sqrt{1+4k}}$$$$1+4k=p^2$$$$4dk=2pdp$$$$\frac{-1}{2}\int \frac{pdp}{2\times (1+\frac{p^2-1}{4})p}$$$$=\frac{-1}{4}\int \frac{4dp}{(p^2+3)}=\frac{-1}{\sqrt{3}}{tan}^{-1}\left(\frac{p}{\sqrt{3}}\right)+c$$$$=\frac{-1}{\sqrt{3}}{tan}^{-1}\left(\frac{\sqrt{1+4k}}{\sqrt{3}}\right)+c$$$$=\frac{-1}{\sqrt{3}}{tan}^{-1}\left(\frac{\sqrt{1+4t^2}}{\sqrt{3}}\right)+c$$$$=\frac{-1}{\sqrt{3}}{tan}^{-1}\left(\frac{\sqrt{1+\frac{4}{x^2}}}{\sqrt{3}}\right)+c$$

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