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Question Number 157442 by bobhans last updated on 23/Oct/21

$$\int \frac{\mathrm{dx}}{\mathrm{x}-\sqrt{{\mathrm{x}}^2+2\mathrm{x}+2}}$$

Answered by MJS_new last updated on 23/Oct/21

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

Answered by cortano last updated on 23/Oct/21

$$\frac{1}{x-\sqrt{x^2+2x+2}}=\frac{x+\sqrt{x^2+2x+2}}{-2x-2}$$$$I=-\int \frac{x}{2x+2}dx-\int \frac{\sqrt{x^2+2x+2}}{2x+2}dx$$$$I_1=-\frac{1}{2}\int \frac{x+1-1}{x+1}dx=-\frac{1}{2}x+\ln \mid x+1\mid +c_1$$$$I_2=-\frac{1}{2}\int \frac{\sqrt{{(x+1)}^2+1}}{x+1}dx$$$$x+1=\tan u$$$$I_2=-\frac{1}{2}\int \frac{\sec u}{\tan u}\sec {}^{2}du$$$$I_2=-\frac{1}{2}\int cscx\sec {}^{2}xdx$$$$IBP\Rightarrow \{\begin{array}{l}u=cscx\Rightarrow du=-cscx\cot xdx\\ v=\tan x\end{array}$$$$I_2=-\frac{1}{2}\{cscx\tan x+\int cscxdx\}$$$$I_2=-\frac{1}{2}cscx\tan x-\frac{1}{2}\ln \mid cscx-\cot x\mid +c_2$$$$I=I_1+I_2$$

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