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Question Number 63116 by mugan deni last updated on 29/Jun/19

$$\int \frac{1+\times }{\sqrt{1+{\times }^2}}dx$$

Commented by mathmax by abdo last updated on 30/Jun/19

$$letusethechangementx=tan\theta \Rightarrow$$$$\int \frac{1+x}{\sqrt{1+x^2}}dx=\int \frac{1+tan\theta }{\sqrt{1+{tan}^2\theta }}(1+{tan}^2\theta )d\theta =\int \sqrt{1+{tan}^2\theta }(1+tan\theta )d\theta$$$$=\int \frac{1+tan\theta }{cos\theta }d\theta =\int \frac{d\theta }{cos\theta }+\int \frac{sin\theta }{{cos}^2\theta }d\theta$$$$\int \frac{sin\theta }{{cos}^2\theta }=\frac{1}{cos\theta }+c_1$$$$\int \frac{d\theta }{cos\theta }{=}_{tan\left(\frac{\theta }{2}\right)=u}\int \frac{2du}{(1+u^2)\frac{1-u^2}{1+u^2}}=\int \frac{2du}{1-u^2}=\int (\frac{1}{1-u}+\frac{1}{1+u})du$$$$=ln\mid \frac{1+u}{1-u}\mid +c_2=ln\mid \frac{1+tan\left(\frac{\theta }{2}\right)}{1-tan\left(\frac{\theta }{2}\right)}\mid +c_2=ln\mid tan(\frac{\pi }{4}+\frac{\theta }{2})\mid \Rightarrow$$$$wehave1+x^2=\frac{1}{{cos}^2\theta }\Rightarrow {cos}^2\theta =\frac{1}{1+x^2}\Rightarrow cos\theta =\frac{1}{\sqrt{1+x^2}}\Rightarrow$$$$\int \frac{1+x}{\sqrt{1+x^2}}dx=\sqrt{1+x^2}+ln\mid tan(\frac{\pi }{4}+\frac{arctanx}{2})\mid +C$$$$anotherway$$$$\int \frac{1+x}{\sqrt{1+x^2}}dx=\int \frac{dx}{\sqrt{1+x^2}}+\int \frac{xdx}{\sqrt{1+x^2}}=ln(x+\sqrt{1+x^2}+\sqrt{1+x^2}+C$$$$andthiswayiseazy...$$

Answered by Hope last updated on 29/Jun/19

$$\int \frac{dx}{\sqrt{1+x^2}}+\int \frac{xdx}{\sqrt{1+x^2}}$$$$x=tanadx={sec}^{2}ada$$$$\int \frac{{sec}^{2}ada}{seca}+\int \frac{tana\times {sec}^{2}ada}{seca}$$$$\int secada+\int tanasecada$$$$ln(seca+tana)+seca+c$$$$ln(x+\sqrt{1+x^2})+\sqrt{1+x^2}+c$$

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