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Evaluate $\displaystyle\int_0^1 \sqrt{x^2+4x+1}\,dx$

I tried simplify by doing this: $$\int_0^1 \sqrt{x^2+4x+1}\:dx=\displaystyle\int_0^1 \sqrt{(x+2)^2-3}\:dx$$ Then, by letting $t=x+2$, $dt=dx$ $$\int_0^1 \sqrt{t^2-3}\:dt=\displaystyle\int_0^1 \sqrt{(t-\sqrt 3)(t+\sqrt 3)}\:dt$$ I tried solving $\displaystyle\int_0^1 \sqrt{t^2-3}\:dt$ by substituting: $$k=t^2-3, dt=\frac{dk}{2\sqrt{|k+3|}}$$ But that leads me nowhere: $$\frac{1}{2}\displaystyle\int_0^1 \frac{\sqrt{k}}{\sqrt{|k+3|}}\:dk$$ Any hints?

Robert Z
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asaf92
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    Use the substitution $t=\sqrt3\cosh(u)$. (If you are unfamiliar with the hyperbolic functions, they have properties very similar to those of their trigonometric counterparts.) Alternatively, $t=\sqrt3(u+1/u)$. –  Apr 07 '16 at 12:14
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    And do not forget to adjust the bounds. – Claude Leibovici Apr 07 '16 at 12:15
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    Sorry, I mean $t=\sqrt3(u+1/u)/2$. –  Apr 07 '16 at 12:20

2 Answers2

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HINT:

Set $x+2=\sqrt3\sec u\implies\sqrt3\tan u=\sqrt{x^2+4x+1}$

$\int\sqrt{(x+2)^2-3}dx=\int\sqrt3\tan u(\sqrt3\sec u\tan u)du=3\int(\sec^3u-\sec u)\ du$

See How to integrate $\sec^3 x \, dx$?

Community
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lab bhattacharjee
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Using the Euler substitution $\sqrt{x^2+4x+1}=t-x$, we obtain $\displaystyle x=\frac{t^2-1}{2(t+2)}$ and $\displaystyle dx=\frac{t^2+4t+1}{2(t+2)^2}dt$,

so $\displaystyle\int_0^1\sqrt{x^2+4x+1}dx=\int_1^{1+\sqrt{6}}\left(t-\frac{t^2-1}{2(t+2)}\right)\frac{t^2+4t+1}{2(t+2)^2}dt=\frac{1}{4}\int_1^{1+\sqrt{6}}\frac{(t^2+4t+1)^2}{(t+2)^3}dt$.

Now let $u=t+2, t=u-2, dt=du$ to get

$\displaystyle\frac{1}{4}\int_3^{3+\sqrt{6}}\frac{(u^2-3)^2}{u^3}du=\frac{1}{4}\int_3^{3+\sqrt{6}}\left(u-\frac{6}{u}+9u^{-3}\right)du=\frac{1}{4}\left[\frac{u^2}{2}-6\ln u-\frac{9}{2}u^{-2}\right]_3^{3+\sqrt{6}}$

$\displaystyle=\frac{3}{4}(\sqrt{6}+1)-\frac{3}{2}\ln\left(\frac{3+\sqrt{6}}{3}\right)-\frac{9}{8}\left(\frac{-6(1+\sqrt{6})}{9(15+6\sqrt{6})}\right)=\frac{3}{2}\sqrt{6}-1-\frac{3}{2}\ln\left(\frac{3+\sqrt{6}}{3}\right)$

user84413
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