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Question Number 41846 by maxmathsup by imad last updated on 13/Aug/18

$$find\int \frac{dx}{\sqrt{1+x^2}+\sqrt{1-x^2}}$$

Answered by MJS last updated on 13/Aug/18

$$\int \frac{dx}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\frac{1}{2}\int \frac{\sqrt{1+x^2}}{x^2}dx-\frac{1}{2}\int \frac{\sqrt{1-x^2}}{x^2}dx=$$$$\left[\mathrm{both}\mathrm{integrals}\mathrm{by}\mathrm{parts}\right]$$$$=\frac{1}{2}(\arcsin x+\mathrm{arcsinh}x)+\frac{\sqrt{1-x^2}-\sqrt{1+x^2}}{2x}+C$$

Commented by math khazana by abdo last updated on 14/Aug/18

$$thankyousir.$$

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