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Question Number 94143 by john santu last updated on 17/May/20

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

Commented by abdomathmax last updated on 17/May/20

$$I=\int \frac{dx}{{(x+\sqrt{1+x^2})}^2}wedothechsngementx=sh\left(t\right)$$$$\Rightarrow I=\int \frac{cht}{(sht+cht)}dt=\int \frac{e^t+e^{-t}}{2{(\frac{e^t+e^{-t}}{2}+\frac{e^t-e^{-t}}{2})}^2}dt$$$$=\frac{1}{2}\int \frac{e^t+e^{-t}}{e^{2t}}dt=\frac{1}{2}\int e^{-2t}(e^t+e^{-t})dt$$$$=\frac{1}{2}\int (e^{-t}+e^{-3t})dt=-\frac{1}{2}{e}^{-t}-\frac{1}{6}{e}^{-3t}+C$$$$t=argsh\left(x\right)=ln(x+\sqrt{1+x^2})\Rightarrow$$$$I=-\frac{1}{2(x+\sqrt{1+x^2})}-\frac{1}{6{(x+\sqrt{1+x^2})}^3}+C$$

Commented by i jagooll last updated on 17/May/20

$$\mathrm{what}\mathrm{the}\mathrm{meaning}\mathrm{cht}?\cosh \left(\mathrm{t}\right)?$$

Commented by mathmax by abdo last updated on 17/May/20

$$cht=\frac{e^t+e^{-t}}{2}(\to hyperboliccos)$$$$sh\left(t\right)=\frac{e^t-e^{-t}}{2}(\to hyperbolicsin)$$

Answered by john santu last updated on 17/May/20

$$\frac{1}{x+\sqrt{1+x^2}}\times \frac{x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}}=\frac{x-\sqrt{1+x^2}}{-1}$$$$\int \frac{dx}{{(x+\sqrt{1+x^2})}^2}=\int {\left(\frac{x-\sqrt{1+x^2}}{-1}\right)}^{2}dx$$$$=\int x^2-2x\sqrt{1+x^2}+1+x^{2}dx$$$$=\int (2x^2+1-2x\sqrt{1+x^2})dx$$$$=\frac{2}{3}{x}^3+x-\frac{2}{3}\sqrt{{(1+x^2)}^3}+c$$

Commented by i jagooll last updated on 17/May/20

cool man ����

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