Question

50 POINTS INTEGRATION PROBLEM

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integrate

$\frac{ {x}^{2 } - 1 }{ {x}^{2} + 1 } \frac{1}{ \sqrt{1 + {x}^{2} } }dx$
Need correct explanation

NO DIRECT answers

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Answer 1

Answer:

Hello mate ,

see the above refer attachment! !

so hope u got your solution! !

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Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\int \dfrac{x^2-1}{x^2+1}\cdot \dfrac{1}{\sqrt{1+x^2}}dx}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int \dfrac{x^2-1}{x^2+1}\cdot \dfrac{1}{\sqrt{1+x^2}}dx=-\dfrac{2x}{\sqrt{x^2+1}}+\ln \left|x+\sqrt{x^2+1}\right|+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\dfrac{x^{2}-1}{x^{2}+1}=-\dfrac{2}{x^{2}+1}+1$

$=\int \left(-\dfrac{2}{x^2+1}+1\right)\dfrac{1}{\sqrt{1+x^2}}dx$

$\text { Expand }\left(-\dfrac{2}{x^{2}+1}+1\right) \dfrac{1}{\sqrt{1+x^{2}}}: \dfrac{x^{2}}{\left(x^{2}+1\right)^{\tfrac{3}{2}}}-\dfrac{1}{\left(x^{2}+1\right)^{\tfrac{3}{2}}}$

$=\int \dfrac{x^2}{\left(x^2+1\right)^{\tfrac{3}{2}}}-\dfrac{1}{\left(x^2+1\right)^{\tfrac{3}{2}}}dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int \dfrac{x^2}{\left(x^2+1\right)^{\tfrac{3}{2}}}dx-\int \dfrac{1}{\left(x^2+1\right)^{\tfrac{3}{2}}}dx$

$\int\dfrac{x^2}{\left(x^2+1\right)^{\tfrac{3}{2}}}dx=-\dfrac{x}{\sqrt{1+x^2}}+\ln \left|x+\sqrt{1+x^2}\right|$

$\int \dfrac{1}{\left(x^2+1\right)^{\tfrac{3}{2}}}dx=\dfrac{x}{\sqrt{1+x^2}}$

$=-\dfrac{x}{\sqrt{1+x^2}}+\ln \left|x+\sqrt{1+x^2}\right|-\dfrac{x}{\sqrt{1+x^2}}$

$=-\dfrac{2x}{\sqrt{x^2+1}}+\ln \left|x+\sqrt{x^2+1}\right|$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$\boxed{\sf{=-\dfrac{2x}{\sqrt{x^2+1}}+\ln \left|x+\sqrt{x^2+1}\right|+C}}$