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in order to solve $\int_{0}^a \sqrt{1+x^2} dx$ someone gave a hint; separating the term into:

(1) $\frac{1}{\sqrt{1+x^2}} + x\cdot \frac{x}{\sqrt{1+x^2}}$

The first one is done. But how to tackle $x\cdot \frac{x}{\sqrt{1+x^2}}$ (without substitution!)? I'd tried integrating by parts, ending with some nonsense. Any hints?

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  • $\begingroup$ Why must you avoid substitution? $\endgroup$
    – Simply Beautiful Art
    Commented Oct 14, 2017 at 11:39
  • $\begingroup$ Simply because of training other ways/methods.. :) $\endgroup$
    – Vazrael
    Commented Oct 14, 2017 at 12:06
  • $\begingroup$ Okay $\ddot\smile$ $\endgroup$
    – Simply Beautiful Art
    Commented Oct 14, 2017 at 12:07

2 Answers 2

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$$I=\int\sqrt{1+x^2}\,dx=\int\dfrac{1}{\sqrt{1+x^2}}\,dx+\int x\dfrac{2x}{2\sqrt{1+x^2}}\,dx$$ By parts $$\int x\dfrac{2x}{2\sqrt{1+x^2}}\,dx = x\sqrt{1+x^2}-\int\sqrt{1+x^2}\,dx = x\sqrt{1+x^2}-I $$ then $$2I=\int\dfrac{1}{\sqrt{1+x^2}}\,dx+x\sqrt{1+x^2}$$

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  • $\begingroup$ cool, thanks! That's really nice. $\endgroup$
    – Vazrael
    Commented Oct 14, 2017 at 12:28
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    $\begingroup$ Your welcome.!. $\endgroup$
    – Nosrati
    Commented Oct 14, 2017 at 12:29
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Integrate by parts to get

$$2\int_0^a\frac{x^2}{\sqrt{1+x^2}}~{\rm d}x=a\sqrt{1+a^2}-\int_0^a\sqrt{1+x^2}~{\rm d}x$$

So now you can relate the two.

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  • $\begingroup$ Thank you too as well, now it makes sense :) $\endgroup$
    – Vazrael
    Commented Oct 14, 2017 at 12:28

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