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Question Number 26769 by julli deswal last updated on 29/Dec/17

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{[x+\sqrt{x^2+1}]}^3}=$$

Commented by prakash jain last updated on 29/Dec/17

$$u=x+\sqrt{x^2+1}$$$$x=0,u=1;x=\mathrm{\infty },u=\mathrm{\infty }$$$$\Rightarrow {(u-x)}^2=x^2+1$$$$\Rightarrow u^2-2ux=1$$$$x=\frac{1}{2}(u-\frac{1}{u})$$$$dx=\frac{1}{2}(1+\frac{1}{u^2})du$$$${\int }_1^{\mathrm{\infty }}\frac{\frac{1}{2}(1+\frac{1}{u^2})du}{u^3}$$$$=\frac{1}{2}{\int }_1^{\mathrm{\infty }}\frac{1}{u^3}du+\frac{1}{2}{\int }_1^{\mathrm{\infty }}\frac{1}{u^5}du$$$$=\frac{1}{2}{[-\frac{1}{2u^2}]}_1^{\mathrm{\infty }}+\frac{1}{2}{[-\frac{1}{4u^4}]}_1^{\mathrm{\infty }}$$$$=\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}$$$$=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$$

Answered by prakash jain last updated on 29/Dec/17

$$\frac{3}{8}\left(seecomment\right)$$

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