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Question Number 36997 by rahul 19 last updated on 07/Jun/18

$$\int \sqrt{\frac{1+x}{x}}dx=?$$

Commented by behi83417@gmail.com last updated on 07/Jun/18

$$x={tg}^{2}t\Rightarrow dx=2tgt(1+{tg}^{2}t)dt$$$$I=\int \sqrt{\frac{1+{tg}^{2}t}{{tg}^{2}t}.}2tgt(1+{tg}^{2}t)dt=$$$$=\int \frac{\frac{1}{cost}}{tgt}.2tgt(1+{tg}^{2}t)dt=2\int {sec}^{3}tdt=$$$$=sect.tgt+ln(sect+tgt)+C=$$$$=\sqrt{x(1+x)}+ln(\sqrt{x}+\sqrt{1+x})+C.◼$$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

$$\int \frac{1+x}{\sqrt{x+x^2}}dx$$$$\frac{1}{2}\int \frac{1+1+2x}{\sqrt{x+x^2}}dx$$$$=\frac{1}{2}\int \frac{dx}{\sqrt{x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}}}+\frac{1}{2}\int \frac{d(x+x^2)}{\sqrt{x+x^2}}$$$$=\frac{1}{2}\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}}+\frac{1}{2}\times \frac{{(x+x^2)}^{\frac{1}{2}}}{\frac{1}{2}}$$$$useformula\int \frac{dx}{\sqrt{X^2-A^2}}$$

Answered by MJS last updated on 07/Jun/18

$$\int \sqrt{\frac{1+x}{x}}dx=\int \frac{\sqrt{1+x}}{\sqrt{x}}dx=$$$$[t=\sqrt{x}\to dx=2\sqrt{x}dt]$$$$=2\int \sqrt{1+t^2}=$$$$[u=\arctan t\to dt={\sec }^{2}udu]$$$$=2\int {\sec }^{2}u\sqrt{1+{\tan }^{2}u}du=2\int {\sec }^{3}udu=$$$$[\int {\sec }^{n}udu=\frac{{\sec }^{n-2}u\tan u}{n-1}+\frac{n-2}{n-1}\int {\sec }^{n-2}udu]$$$$=\sec u\tan u+\int \sec udu=$$$$=\sec u\tan u+\ln (\sec u+\tan u)=$$$$=t\sqrt{1+t^2}+\ln (\sqrt{1+t^2}+t)=$$$$=\sqrt{x(1+x)}+\ln (\sqrt{1+x}+\sqrt{x})+C$$

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