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I need a help solving this integration equation,$$\int {y \over \sqrt{1+y^2}} dy$$

I am quite weak with maths. I would really appreciate if someone would guide me solving this.

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Try computing the derivative of $f(x) = \sqrt{1+x^2}$. –  copper.hat Jan 8 '13 at 3:27
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3 Answers 3

up vote 5 down vote accepted

Let $u=1+y^2$.

Then:

$du=2y\,dy\implies {1\over 2}\,du=y\,dy$

And:

$$\displaystyle{\int{y\over \sqrt{1+y^2}}\,dy={1\over 2}\int {du\over \sqrt u}= {1\over 2}\int u^{-1/2}\,du={1\over 2}{u^{1/2}\over 1/2}+C=\sqrt{1+y^2}+C.}$$

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kudos! I wish i could think this stuff –  Kishan Thobhani Jan 8 '13 at 3:40
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First check that

$$\int\frac{f'(x)\,dx}{\sqrt{f(x)}}=2\sqrt{f(x)}+C\,\,,\,\,C=\text{ a constant}$$

and now just observe that

$$y=\frac{1}{2}2y=\frac{1}{2}\frac{d}{dy}(1+y^2)=\frac{1}{2}(1+y^2)'$$

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You may also substitute $y=\tan(\theta)$, and $dy=\sec^2(\theta)d\theta$. $$\int \frac{y}{\sqrt{1+y^2}} dy=\int \frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}\sec^2(\theta)d\theta=\int \tan(\theta)\sec(\theta)d\theta=\sec(\theta)$$ $$\sec(\theta)=\sqrt{1+\tan^2(\theta)}=\sqrt{1+y^2}$$ So the answer is $\sqrt{1+y^2}+C$.

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