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Question Number 42791 by maxmathsup by imad last updated on 02/Sep/18

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

Answered by MJS last updated on 02/Sep/18

$$\mathrm{I}\mathrm{know}\mathrm{this}\mathrm{because}\mathrm{of}\mathrm{the}\mathrm{hyperbola}\mathrm{formuls}$$$$y=\sqrt{x^2+1}$$$$y^{\prime }=\frac{x}{\sqrt{x^2+1}}$$$$y^{″}=\frac{1}{{(x^2+1)}^{\frac{3}{2}}}$$$$\Rightarrow \int \frac{dx}{(x^2+1)\sqrt{x^2+1}}=\int \frac{dx}{{(x^2+1)}^{\frac{3}{2}}}=\frac{x}{\sqrt{x^2+1}}+C$$

Commented by math khazana by abdo last updated on 03/Sep/18

$$thankyousir.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

$$x=tan\alpha dx={sec}^2\alpha d\alpha$$$$\int \frac{{sec}^2\alpha d\alpha }{{\left({sec}^2\alpha \right)}^{\frac{3}{2}}}$$$$\int cos\alpha d\alpha$$$$sin\alpha +c$$$$=\frac{x}{\sqrt{1+x^2}}+c$$

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