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Question Number 183845 by Michaelfaraday last updated on 30/Dec/22

Answered by MJS_new last updated on 31/Dec/22

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

Answered by MJS_new last updated on 31/Dec/22

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

Commented by Michaelfaraday last updated on 31/Dec/22

$$thankssir$$

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