problematic integral #2

**Shock** · March 21st 2010, 02:12 AM

Hi.
Evaluate:

$\int <br /> {\sqrt{x^2+x+3} dx}$

**mr fantastic** · March 21st 2010, 02:15 AM

Originally Posted by Shock

Hi.
Evaluate:

$\int <br /> {\sqrt{x^2+x+3} dx}$

See here: integrate Sqrt[x^2 + x + 3] - Wolfram|Alpha

(Click on Show steps).

**Moo** · March 21st 2010, 02:54 AM

Hello,

By completing the square, we have $x^2+x+3=\left(x+\tfrac 12\right)^2+\tfrac{11}{4}=\tfrac{11}{4}\left(\tfra c{4}{11}\cdot\left(x+\tfrac 12\right)^2+1\right)$

Using the usual substitutions, let $\sinh(u)=\sqrt{\tfrac{4}{11}}\cdot\left(x+\tfrac 12\right)$

(thus $\cosh(u) ~du=\sqrt{\tfrac{4}{11}} ~dx$ )

And you'll have $\int \sqrt{\tfrac{11}{4}}\cdot\sqrt{\tfrac{11}{4}} \cdot \sqrt{\sinh^2(u)+1}\cdot\cosh(u) ~du=\tfrac{11}{4}\int \cosh^2(u) ~du$

But $2\cosh^2(u)=1+\cosh(2u)$

So $\tfrac{11}{4}\int \cosh^2(u) ~du=\tfrac{11}{8}\left(x+\tfrac 12 \cdot \sinh(2u)\right)+C$

Now, back substitute

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