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Question Number 145775 by Engr_Jidda last updated on 08/Jul/21

$$\int \frac{2x+1}{\sqrt{x^2+4x+5}}dx$$

Answered by mathmax by abdo last updated on 08/Jul/21

$$\mathrm{{\rm Y}}=\int \frac{2\mathrm{x}+4-3}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+5}}\mathrm{dx}=\int \frac{2\mathrm{x}+4}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+5}}\mathrm{dx}-3\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+5}}$$$$=2\sqrt{{\mathrm{x}}^2+4\mathrm{x}+5}-3\mathrm{I}$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+4+1}}=\int \frac{\mathrm{dx}}{\sqrt{{(\mathrm{x}+2)}^2+1}}$$$${=}_{\mathrm{x}+2=\mathrm{sht}}\int \frac{\mathrm{cht}}{\mathrm{cht}}\mathrm{dt}=\mathrm{t}+\mathrm{K}=\mathrm{argsh}(\mathrm{x}+2)+\mathrm{K}$$$$=\log (\mathrm{x}+2+\sqrt{1+{(\mathrm{x}+2)}^2})+\mathrm{K}\Rightarrow$$$$\mathrm{{\rm Y}}=2\sqrt{{\mathrm{x}}^2+4\mathrm{x}+5}-3\log (\mathrm{x}+2+\sqrt{{\mathrm{x}}^2+4\mathrm{x}+5})+\mathrm{K}$$

Answered by ArielVyny last updated on 08/Jul/21

$$\int \frac{2x+1}{\sqrt{{(x+2)}^2+1}}dx$$$$t=x+2\to dt=dx$$$$\int \frac{2(t-2)+1}{\sqrt{t^2+1}}dt=\int \frac{2t}{\sqrt{t^2+1}}dt-3\int \frac{1}{\sqrt{t^2+1}}dt$$$$2\sqrt{t^2+1}-3argsh\left(t\right)+cte$$

Commented by ArielVyny last updated on 08/Jul/21

$$2\sqrt{x^2+2x+5}-3argsh(x+2)+cte$$

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