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Question Number 166570 by cortano1 last updated on 22/Feb/22

Answered by bobhans last updated on 22/Feb/22

$$\mathrm{I}=\int \frac{{(\mathrm{x}+1)}^2+2}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}=\int \frac{\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}+\int \frac{2}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}$$$${\mathrm{I}}_1=\frac{1}{2}\int \frac{\mathrm{d}({\mathrm{x}}^2+1)}{\sqrt{{\mathrm{x}}^2+1}}=\sqrt{{\mathrm{x}}^2+1}+{\mathrm{c}}_1$$$${\mathrm{I}}_2=\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+1}},[\mathrm{x}=\tan \theta ]$$$${\mathrm{I}}_2=\int \frac{\sec {}^2\theta }{\sec \theta }\mathrm{d}\theta =\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2+1}\mid +{\mathrm{c}}_2$$$${\mathrm{I}}_3=\int \frac{2}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx},[\mathrm{x}=\tan \mathrm{y}]$$$${\mathrm{I}}_3=2\int \frac{\sec {}^2\mathrm{y}}{(1+\tan \mathrm{y})\sec \mathrm{y}}\mathrm{dy}=2\int \frac{\sec \mathrm{y}}{1+\tan \mathrm{y}}\mathrm{dy}$$$${\mathrm{I}}_3=2\int \frac{\mathrm{dy}}{\cos \mathrm{y}+\sin \mathrm{y}}=2\int \frac{\mathrm{dy}}{\sqrt{2}\cos (\mathrm{y}-45°)}$$$${\mathrm{I}}_3=\sqrt{2}\int \sec (\mathrm{y}-45°)\mathrm{dy}=\sqrt{2}\ln \mid \frac{1+\sin (\mathrm{y}-45°)}{\cos (\mathrm{y}-45°)}\mid +{\mathrm{c}}_3$$$${\mathrm{I}}_3=\sqrt{2}\ln \mid \frac{\sqrt{2}(1+\frac{1}{\sqrt{2}}\sin \mathrm{y}-\frac{1}{\sqrt{2}}\cos \mathrm{y})}{\cos \mathrm{y}+\sin \mathrm{y}}\mid +{\mathrm{c}}_3$$$${\mathrm{I}}_3=\sqrt{2}\ln \mid \frac{\sqrt{2}+\sin \mathrm{y}-\cos \mathrm{y}}{\cos \mathrm{y}+\sin \mathrm{y}}\mid +{\mathrm{c}}_3$$

Commented by peter frank last updated on 22/Feb/22

$$\mathrm{great}$$

Commented by MJS_new last updated on 23/Feb/22

$$I_3=2\int \frac{dx}{(x+1)\sqrt{x^2+1}}=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{\sqrt{x^2+1}}{t}dt]$$$$=4\int \frac{dt}{t^2+2t-1}=\sqrt{2}\int (\frac{1}{t+1-\sqrt{2}}-\frac{1}{t+1+\sqrt{2}})dt=$$$$=\sqrt{2}\ln \frac{t+1-\sqrt{2}}{t+1+\sqrt{2}}=$$$$=\sqrt{2}\ln \mid \frac{x-1+\sqrt{2(x^2+1)}}{x+1}\mid +C$$$$\mathrm{but}\mathrm{of}\mathrm{course}\mathrm{you}\mathrm{get}\mathrm{the}\mathrm{same}:$$$$\frac{\sqrt{2}+\sin y-\cos y}{\cos y+\sin y}=$$$$[y=\arctan x]$$$$=\frac{x-1+\sqrt{2(x^2+1)}}{x+1}$$

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