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Question Number 36595 by rahul 19 last updated on 03/Jun/18

$$\int \frac{xdx}{\sqrt{1+x^2+\sqrt{{(1+x^2)}^3}}}=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$t=1+x^2$$$$dt=2xdx$$$$\int \frac{1}{2}\times \frac{dt}{\sqrt{t+\sqrt{t^3}}}$$$$\frac{1}{2}\int \frac{dt}{\sqrt{t+t\sqrt{t}}}$$$$\frac{1}{2}\int \frac{dt}{\sqrt{t}(\sqrt{1+\sqrt{t})}}$$$$k^2=1+\sqrt{t}$$$$2kdk=0+\frac{1}{2\sqrt{t}}dt$$$$4kdk=\frac{dt}{\sqrt{t}}$$$$\frac{1}{2}\int \frac{4kdk}{k}$$$$2k+c$$$$2(1+\sqrt{t})+c$$$$2(1+\sqrt{1+x^2})+c....ans$$

Commented by rahul 19 last updated on 03/Jun/18

Typo in 2nd last line .

Answered by Joel579 last updated on 03/Jun/18

$$I=\int \frac{x}{\sqrt{(1+x^2)+{(1+x^2)}^{3/2}}}dx(u^2=1+x^2\to 2udu=2xdx)$$$$=\int \frac{x}{\sqrt{u^2+u^3}}.\frac{2udu}{2x}$$$$=\int \frac{udu}{\sqrt{u^2(1+u)}}$$$$=\int \frac{du}{\sqrt{1+u}}$$$$=2\sqrt{1+u}+C$$$$=2\sqrt{1+\sqrt{1+x^2}}+C$$

Commented by rahul 19 last updated on 03/Jun/18

thanks sir����

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